User michael blackmon - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T05:32:18Z http://mathoverflow.net/feeds/user/8843 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53929/locales-and-topology Locales and Topology. Michael Blackmon 2011-01-31T23:05:50Z 2013-01-09T05:00:19Z <p>As someone more used to point-set topology, who is unfamiliar with the inner workings of lattice theory, I am looking to learn about the localic interpretation of topology, of which I only have a limited understanding. As such, I have some questions:</p> <ol> <li>What are some accessible texts or online references on the subject?</li> <li>What are some recent results in point-free topology that are unique to the subject, i.e., not translations of results from general topology into localic language? </li> </ol> http://mathoverflow.net/questions/113805/a-classical-riddle-answered A Classical Riddle Answered Michael Blackmon 2012-11-19T07:26:22Z 2012-11-19T13:20:24Z <p>For a model M and object $\theta$, we may define the two predicates</p> <p>$$\mathcal{C}(M, \theta ) \iff M \vDash \forall \alpha\ \forall f: \alpha \to \dot{\theta}\ \exists \gamma \in \dot{\theta}\ \forall \xi \in \alpha\ (f(\xi) \neq \gamma)$$</p> <p>(there is nothing that can be brought into one-to-one correspondence with $\theta$ with in $M$)</p> <p>$$\mathbb{C}(M, \theta) \iff M \vDash \forall f\in C(\mathbb{R})\ \exists O \subset \mathbb{R} ( O \triangle f[\theta]\ \text{ is meager})$$</p> <p>(the image of $\theta$ under every continuous function with in M, has meager difference with an open set.)</p> <p>Now, here comes the question: Can these two statements every be compared?</p> <ul> <li><p>It turns out the answer depends strongly on the particular context in which the statement or question was asked. Formally these statements have no meaning without an interpretation which witnesses the consistency of the principles which govern the interpretation. As statements of language they are just symbols and words arranged so that once we give them context they have meaning.</p></li> <li><p>In this particular situation there is no simple correct answer, because just like the emotions of the people that will read this particular paragraph and the apology this mathematician is humbly attempting to convey to another mathematician, it is impossible to separate regularity from cardinality without first asserting that for some rational $\delta$ and ordinal $\omega$, we have $\delta &lt; \omega$ or $\omega &lt; \delta$.</p> <p>In much the same way an apology is different from the emotions it illicits, the word emotion is hollow without an instance of an apology to interpret it. As such these two statements and types of objects must always remain incompatible; because otherwise they will inevitably lead to a contradiction. </p></li> </ul> <p>-- MB.</p> http://mathoverflow.net/questions/112466/can-assumptions-about-forcing-produce-mice Can Assumptions about forcing produce Mice? Michael Blackmon 2012-11-15T08:56:16Z 2012-11-19T07:42:58Z <p>This is going to take some build up to completely describe what is a very strange question I seem to have walked into by accident: </p> <p>For every partial order $\mathbb{P}$ and <strong>regular</strong> cardinal $\lambda > \omega$ we can define the following two statements </p> <p>$$\mathcal{C}(\mathbb{P}, \lambda) \iff 1 \Vdash_{\mathbb{P}} \forall \alpha \in \check{\lambda}\ \forall f: \alpha \to \check{\lambda}\ \exists \gamma \in \check{\lambda}\ \forall \xi \in \alpha\ (f(\xi) \neq \gamma)$$</p> <p>(this is the formalized version of the statement "$\mathbb{P}$ preserves $\lambda$ is a cardinal" in the forcing language, this statement is normally certified by reasoning which does not involve the forcing relation and depends on the structure of $\mathbb{P}$-names) </p> <p>and</p> <p>$$Cof(\mathbb{P}, \lambda) \iff 1 \Vdash_{\mathbb{P}} \forall \alpha \in \check{\lambda}\ \forall f:\alpha \to \check{\lambda}\ \exists \gamma \in \check{\lambda} \ (\sup(ran(f)) \le \gamma)$$</p> <p>(Again a forcing language version of the statement $\mathbb{P}$ preserves $\forall \alpha \in \lambda \ (cf(\alpha) &lt; cf(\lambda))$: we had to be careful here because we need to be able to distinguish between the two (If this is not the correct way to formalize this please let me know.))</p> <p>Now, here comes the question: Does the following conjunction:</p> <p>$\exists \lambda > \omega\ \exists\ \mathbb{P}$ such that</p> <ul> <li>$\lambda$ is a <strong>Regular</strong> cardinal.</li> <li>$\vert \mathbb{P} \vert = \lambda^{+}$</li> <li>$\forall \mu \ (\mu$ is a cardinal $\implies \mathcal{C}(\mu,\mathbb{P}))$ </li> <li>$\neg Cof(\lambda, \mathbb{P})$</li> </ul> <p>Imply there is an inner model with a measurable cardinal? (changed based on the answers.)</p> <p>(Namba for $\omega_2$ and threading a generic square collapse cardinals; moreover if $0^\sharp$ exists then $\aleph_\omega^{V}$ is regular in $L$ producing a model in some sense)</p> <p>Edit:</p> <p><strong>(It was not my intention to scare a lot of nice mice)</strong></p> <p><strong>(also, mice need to be more damn direct and stop subtly hinting things.... didn't realize what was going on until just now....)</strong></p> http://mathoverflow.net/questions/91027/mathfrakc-universal-linear-order $\mathfrak{c}$-universal linear order Michael Blackmon 2012-03-12T22:53:13Z 2012-05-02T11:08:02Z <p>I've been told once or twice that the following holds:</p> <blockquote> <p>There is a model of $ZFC+MA+\neg CH$ in which there is a $\mathfrak{c}$-universal linear order embedded in $(\omega^\omega, \le^\ast)$ </p> </blockquote> <p>Moreover, several people have attributed the construction of such a model to Hugh Woodin. The context that seems the most natural for this construction to appear is in his work on automatic-continuity, however for the life of me, I can't seem to find it.</p> <p>It might also be the case that the embedding is into $\mathcal{P}(\omega)/fin$.</p> <p>Does anybody know of a reference for this? Or if Woodin did indeed construct it (I only ask because my searches have yielded little fruit)?</p> <p>Edit: </p> <p>In this context a $\mathfrak{c}$-universal linear order $(L, \prec)$ has the following property: $\vert L \vert \le \mathfrak{c}$ and for every linear order $(\ell, &lt;)$, with $\vert \ell \vert \le \mathfrak{c}$ there is an embedding $\varphi_\ell:\ell \rightarrow L$ respecting the linear order of $\ell$.</p> <hr> <p>Thought I should share what I can find/know:</p> <p>So far I've been able to find models for the following</p> <blockquote> <p>$ZFC+ \neg CH$ and there is such a $\mathfrak{c}$-universal linear order in $(\omega^\omega,\le^\ast)$.</p> </blockquote> <p>This is due to Laver, <a href="http://www.sciencedirect.com/science/article/pii/S0049237X08716306" rel="nofollow">http://www.sciencedirect.com/science/article/pii/S0049237X08716306</a></p> <p>In addition, there is also </p> <blockquote> <p>$ZFC+MA+\neg CH$ and no such order is embedded in $(\omega^\omega,\le^\ast)$</p> </blockquote> <p>There seem to be several models for this one, Laver cites two, one from Solovay and one from Kunen. I can only assume the Solovay model is the same constructed in Theorem 5.7 (p. 201, "Hausdroff Gaps and Limits", by Frankiewicz and Zbierski)</p> <p>In addition there is</p> <blockquote> <p>$CH\implies$ $(\omega^\omega,\le^\ast)$ contains a $\mathfrak{c}$-universal linear order</p> </blockquote> <p>Which should be contained in (or atleast hinted at in) "Model Theory", by Chang and Keisler. If not there, the lack of $(\omega,\omega)$-gaps in $(\omega^\omega,\le^\ast)$ should produce it rather quickly.</p> <p>So yeah, it would seem that the only link missing in this puzzle is the one I can't find the reference for.</p> http://mathoverflow.net/questions/90862/what-are-some-examples-of-ingenious-unexpected-constructions/90883#90883 Answer by Michael Blackmon for What are some examples of ingenious, unexpected constructions? Michael Blackmon 2012-03-11T07:02:19Z 2012-03-11T07:18:33Z <p>The best one I can think of would be: The Attempts to Resolve The Continuum Hypothesis, and The Souslin Hypothesis.</p> <p>The list of tools developed to deal with the new and subtle problems that cropped up while trying to either affirm or refute these two classical problems is almost unending. The main take away from all of this work was that the universe is much more flexible than we had ever dreamed. </p> <p>A short list:</p> <ol> <li>Forcing ( <a href="http://en.wikipedia.org/wiki/Forcing_%28mathematics%29" rel="nofollow">wiki</a> ) <ol> <li><a href="http://en.wikipedia.org/wiki/List_of_forcing_notions" rel="nofollow">http://en.wikipedia.org/wiki/List_of_forcing_notions</a></li> </ol></li> <li>Model Theory ( <a href="http://en.wikipedia.org/wiki/Model_theory" rel="nofollow">wiki model theory</a> )</li> <li>Inner Model Theory ( <a href="http://en.wikipedia.org/wiki/Inner_model_theory" rel="nofollow">inner model theory</a> ) <ol> <li><a href="http://en.wikipedia.org/wiki/Constructible_universe" rel="nofollow">The Original Inner Model</a></li> </ol></li> <li>The Various Large Cardinal Hypothesis which have been developed ( <a href="http://cantorsattic.info/Cantor%2527s_Attic" rel="nofollow">cantors attic</a> ) and Kunen's delimiting result ( <a href="http://en.wikipedia.org/wiki/Kunen%27s_inconsistency_theorem" rel="nofollow">Kunen's inconsistency theorem</a> )</li> </ol> http://mathoverflow.net/questions/90820/set-theories-without-junk-theorems/90882#90882 Answer by Michael Blackmon for Set theories without "junk" theorems? Michael Blackmon 2012-03-11T06:09:07Z 2012-03-11T07:08:34Z <p>Among the many subtle realities of mathematics in the 21st century, the most amazing is the lack of imagination. The language of set theory is built from the ground up to be as simple as possible. To appreciate the complexity inherent and information encoded in such simple statements (even the ones you might not find aesthetically pleasing) requires detachment. </p> <p>This detachment I'm talking about is the clear distinction between: syntax and semantics. Statements made in the formal language have absolutely no meaning outside of formal manipulation, and so are not meant to be seen as anything more than symbols without meaning.</p> <p>It is only when you attach meaning (or an interpretation) to these symbols that something of value can be said. </p> <p><strong>That having been said:</strong></p> <p>The examples you give are not actually statements in the language of set theory; they are artifacts of a general lack of communication between logic/model theory and the rest of mathematics. The symbols you strung together (1, $2$, 5, $4 \subset 54$, $\cap$, and so on) are examples of defined notions, which are used as a convenience.</p> <p>And when we attach meaning to these statements something amazing happens:</p> <p>What was $2 \in 3$ becomes the obviously true</p> <blockquote> <p>$\{ \{\}, \{\{\}\} \} \in \{\{\}, \{ \{\}, \{\{\}\} \}\}$</p> </blockquote> <p>and $1 \in \langle 0, 3 \rangle$ becomes</p> <blockquote> <p>$\{\{\}\} \in \{ \{ \{\} \}, \{ \{\{\}, \{ \{\}, \{\{\}\} \}\} \}\}$</p> </blockquote> <p><strong>In Summary:</strong></p> <p>You are confusing the formal language with the actual interpretation of the language. </p> <p>As such you are faced with something every body has known since the 19th century: </p> <p>Our perception imposes "phantom" structure on the universe in an attempt to have it make sense; not the other way around.</p> <p>PS: Feel free to edit. You also might want to change the title, since the post I wanted to put here would have gotten me banned.</p> http://mathoverflow.net/questions/76995/why-are-some-axioms-preserved-in-generic-extensions/77026#77026 Answer by Michael Blackmon for Why are some axioms preserved in generic extensions? Michael Blackmon 2011-10-03T09:34:35Z 2011-10-08T01:25:18Z <p><strong>Forward:</strong> This answer does not address the main question "which statements are preserved for every generic extension." Which I suspect has no simple answer.</p> <p><strong>Lets start</strong> by defining what this answer is meant to address: Preservation results fall into two types (modulo fine details):</p> <ol> <li><p>Large structural preservation. An example of this kind is Shoenfield's classic result concerning the absoluteness of $\Sigma^1_2$ statements. These generally take the form: For some statement(s) $\varphi$ and "canonical" inner model $M$, we can show that $ZFC \vdash$"$\varphi$ is absolute for $M,V$ ". This is rapidly transformed into a forcing absoluteness result by cleverly setting up the notion of "canonical" so as to guarantee: if $V \subset N$ are models of $ZFC$ and $V\vDash M$ is "canonical", then $N \vDash M$ is "canonical." (In the case of Shoenfield's result this last bit is trivial since $V\subset N \implies L^V = L^N$.)</p></li> <li><p>Small combinatorial preservation. These results are numerous and deal with individual statements assumed to hold in the ground model and then using this assumption produce that the statement is true in the extension (using either the definability of forcing or constructing explicit names of objects in the extension, or a combination of both.) Moreover, results of this form can be considered as the reason a certain statement holds in an extension. These types of proofs can be understood as being essentially internal to $V$.</p></li> </ol> <p>The contents of this answer will attempt to illustrate two examples of the second method and explain why $AC$ is preserved when passing to any generic extension.</p> <p><strong>Now for a horrible and vague answer to a strange question:</strong> What does using a partial order $\mathbb{P}$ to force over a model $V$ actually do? It lets people with imaginations strictly confined to $V$ imagine/build an exterior universe $V[G]$ (whose properties are bound to $\mathbb{P}$, the forcing relation $\Vdash_{\mathbb{P}}$, and $V$.)</p> <p>From this perspective the original question can be recast as: How does one prove certain statements are preserved by a particular $\mathbb{P}$? The answer: it depends, but in general there are two methods which appear most. From here we will consider a couple examples.</p> <p><strong>Consider the case of $AC$</strong>: Now, $AC$ is equivalent to the statement $\forall X\ \exists \alpha \in ON\,\ f\subset X \times \alpha\ (\ f$ is injective$)$ and we want to show that this statement is preserved between forcing extensions. </p> <p><strong>First Method:</strong> (Brute force combinatorics) (paraphrasing what is found in Kunens' text)</p> <p>We must show $\forall p\in \mathbb{P}\ \exists q \le p\ [q \Vdash \forall \dot{X}\ \exists\check{\alpha}\in ON,\ \dot{f} \subset \dot{X} \times \check{\alpha}\ (\dot{f}$ is an injection with domain $\dot{X})]$. Using the standard facts about the forcing relation $\Vdash$, this becomes: for every $\dot{X} \in V^{\mathbb{P}}$ and $p \in \mathbb{P}$, there exists some $q \le p$, $\alpha \in ON$, and $\dot{f}\in V^{\mathbb{P}}$ such that $q \Vdash [\ \dot{f}$ is an injection with domain $\dot{X}\ ].$</p> <p>In order to show this assertion, we are going to explicitly build the name $\dot{f}$, and produce the condition $q$ which forces "$\dot{f}$ <em>is an injection with domain</em> $\dot{X}$" only using statements $ZF+AC$ can prove.</p> <p>To this end, note that since we are assuming $AC$, we may assume $\dot{X} = \bigcup_{\gamma\in\mu}\{ \langle \sigma_\gamma, q \rangle: q \in A_\gamma \}$, where each $\sigma_\gamma$ is a $\mathbb{P}$-name, $\mu$ is some cardinal, and $A_\gamma$ is an anti-chain such that $\forall q \le p$: we have $q \Vdash [\sigma_\gamma \in \dot{X}]$, if and only if $\{ s \in A_\gamma: s \not\perp q \}$ is maximal below $q$.</p> <p>Let $\dot{g}= \bigcup_{\alpha\in\mu} \{ \langle \rho_\alpha(q), q \rangle: q \in A_\alpha \}$ where $\rho_\alpha(q) = \{\langle \sigma_\alpha, q \rangle, \langle \{\langle \sigma_\alpha, q \rangle, \langle \check{\alpha}, q \rangle, \}, q \rangle \}$. Then, taking $q=p$, $\alpha = \mu$ and $\dot{f} = \dot{g}$ establishes the result.</p> <p><strong>Second Method:</strong> (Sage like and Model Theoretic)</p> <p>The statement "$f$ <em>is an injection</em>" is $\Delta_0$. Moreover, if $f$ is an injection, then $ZF(?C)$ proves that the same holds for all of its subsets. Noting that for every $\dot{X} \in V^{\mathbb{P}}$ there exists some $Y \in V$ such that $1 \Vdash \exists \dot{g} \subset \dot{X} \times \check{Y}\ (\dot{g}$ is an injection $)$ (namely $Y = \{\langle \sigma, q \rangle^\check{\ }: \langle \sigma, q \rangle \in \dot{X} \}$) and by $AC$ there exists some injection $f : Y \to \mu$ (some cardinal $\mu$), the result follows. </p> <p><strong>Conclusion for $AC$:</strong> The reason $AC$ is preserved when passing to a generic extension is essentially because it is equivalent to a "well-positioned" statement. Where "well-positioned" in this case means: it asserts the existence of an object with a $\Delta_0$ property and this $\Delta_0$ property is maintained by each of its subsets, all of which can be turned into a witness for some particular instance of $AC$. In this way, the truth or falsity of $AC$ in the extension does not depend on $\mathbb{P}$ and depends only on whether or not it held in $V$.</p> <p>To contrast this with other weak forms of $AC$: consider $DC_\omega$. In order for $DC_\omega$ to be preserved we must avoid something. In particular we must avoid adding some $R \subset X \times X$ which is an entire binary relation and witnesses the failure of $DC_\omega$ (i.e. there is no $\{x_n:n\in\omega\}$ so that $\langle x_n, x_{n+1} \rangle \in R$) and we have no reason to expect this is the case without knowing something about $\mathbb{P}$.</p> http://mathoverflow.net/questions/77043/a-result-of-shelah-about-the-nonstationary-ideal/77045#77045 Answer by Michael Blackmon for A result of Shelah about the nonstationary ideal Michael Blackmon 2011-10-03T14:12:15Z 2011-10-03T14:12:15Z <p>Jech (in the Chapter "Stationary Sets", from the Handbook of Set Theory) lists the reference as</p> <blockquote> <p>Saharon Shelah. Iterated forcing and normal ideals on $\omega_1$. Israel Journal of Mathematics, 60(3):345–380, 1987.</p> </blockquote> http://mathoverflow.net/questions/75836/the-set-theoretic-multiverse-as-a-bicategory/76343#76343 Answer by Michael Blackmon for The set-theoretic multiverse as a (bi)category Michael Blackmon 2011-09-25T16:27:34Z 2011-09-27T20:06:17Z <p><strong>To Begin:</strong> There are two notions we need to clearly distinguish here: first, the principles which are suspected (or asserted) to hold in the multiverse; and second, the actual intended interpretations of these principles and the implicit bounds placed on them by living in a particular $V$.</p> <p>First, the principles he discusses are first order, which means they live in the world of math proper. Moreover, they are nice, and in fact there is a rather nice model for them. However, this model and any other such model, can in no way reflect anything other then the first-order principles which are asserted to hold in the multiverse. They are simply objects which exhibit the consistency/coherence of said principles (kind of like exhibiting $\{0\}$ with the operation $\{\langle\langle 0, 0 \rangle,0\rangle \}$ and noting that it satisfies the axioms of a group.)</p> <p>Second, under the intended interpretation, there can be no actual <em>object</em> which is <em>the</em> multiverse. This follows directly from the Forcing Extension Axiom. The reason being: it is impossible to internally "close off" a forcing notion, because by asserting a particular generic exists, you have just defined how to get around it. More succinctly put, for any separative $\mathbb{P}$, if $G$ is $\mathbb{P}$-generic over $V$, then $G$ is not $\mathbb{P}$-generic over $V[G]$ (since $1 \Vdash \forall \dot{p}\in\check{\mathbb{P}} \,\exists\dot q\in (\check{\mathbb{P}}\backslash \dot{G})(\dot{q}\le\dot{p})$.) Moreover, the "Absorption into L" and "Countability Principle" combine to imply that any $V$ which thinks it has captured the multiverse, is only lying to itself.</p> <p><strong>My Main Point:</strong> The proper multiverse is a flat out, meta object, in the strongest sense possible. The reason for this: you officially cannot get out in front of it, or out run the strength of its intended interpretation (like you can do with inaccessible cardinals and ZFC.)</p> <p><strong>Addendum:</strong> see comments.</p> <p><strong>Note:</strong> If there is an issue please let me know in the comments</p> http://mathoverflow.net/questions/71201/are-there-examples-of-statements-that-have-been-proven-whose-consistency-proofs-c/71267#71267 Answer by Michael Blackmon for Are there examples of statements that have been proven whose consistency proofs came before their proofs? Michael Blackmon 2011-07-25T22:27:58Z 2011-07-25T22:27:58Z <p>Considering all the answers so far, I thought I might as well add one with a more topological flavor to it</p> <ol> <li><p>The existence of an L-space was known to be consistent for years (See The Handbook of Set-Theoretic Topology, Chapter 7, pg. 295). It was only recently that a ZFC construction was given <a href="http://www.math.cornell.edu/~justin/Ftp/Lspace.pdf" rel="nofollow">here</a></p></li> <li><p>Some of the existence proofs for certain types of embeddings, and automorphisms between Boolean algebras have this flair to them, (See "The fourth head of $\beta\mathbb{N}$" by Ilijas Farah, in Open Problems in Topology II. pg 139.)</p></li> <li><p>Certain types of gnarly questions about coverings of $\mathbb{R}$ involving the forward and inverse images of $\aleph_1$ many continuous functions, have had some success with this <a href="http://mathoverflow.net/questions/53711/cn-and-forcing-reading-a-recent-paper-by-kunen/53728#53728" rel="nofollow">see this answer</a></p></li> </ol> http://mathoverflow.net/questions/65381/m-a-d-family-question/65397#65397 Answer by Michael Blackmon for M.A.D family question Michael Blackmon 2011-05-19T04:28:39Z 2011-05-19T04:28:39Z <p>The article:</p> <p>"Mad families and their neighbors" (2007) by Andreas Blass, Tapani Hyttinen, Yi Zhang </p> <p>(pdf available here: <a href="http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.107.3300&amp;rep=rep1&amp;type=pdf" rel="nofollow">http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.107.3300&amp;rep=rep1&amp;type=pdf</a>)</p> <p>talks about several related notions of MAD, and might be a decent starting point. </p> <p>This is an intriguing construction. In passing I would like to point out that $\kappa^\omega = \kappa \implies \lambda = \kappa$, says that under $GCH$ (so in $L$ for example) the only places the assertion can fail is at a singular cardinal of countable cofinality. As such you might want to have a look at what happens at $\omega_\omega$ under $CH$ (where you have $\forall n > 0 ({\omega_n}^{\omega} = \omega_n)$.) Or more to the point, what happens when you have a sequence of cardinals $\kappa_n : n\in \omega$, such that for each $n\in\omega$, $\lambda_n = \kappa_n$, and $\kappa = \sup\{ \kappa_n: n\in\omega \}$.</p> http://mathoverflow.net/questions/53188/decent-texts-on-categorical-logic Decent Texts on Categorical Logic Michael Blackmon 2011-01-25T04:41:32Z 2011-03-19T18:30:38Z <p>Recently I read the chapter "Doctrines in Categorical Logic" by Kock, and Reyes in the Handbook of Mathematical Logic. And I was quite impressed with the entire chapter. However it is very short, and considering that this copy was published in 1977, possibly a bit out of date. </p> <p>My curiosity has been sparked (especially given the possibilities described in the aforementioned chapter) and would like to know of some more modern texts, and articles written in this same style (that is to say coming from a logic point of view with a strong emphasis on analogies with normal mathematical logic.)</p> <p>However, that being said, <em>I am stubborn as hell, and game for anything.</em></p> <p>So, Recommendations?</p> <p>Side Note: More interested in the preservation of structure, and the production of models than with any sort of proposed foundational paradigm </p> http://mathoverflow.net/questions/7103/topological-rings/53145#53145 Answer by Michael Blackmon for Topological Rings Michael Blackmon 2011-01-24T22:26:20Z 2011-02-14T12:13:20Z <p>While looking through old questions, I came accross this one, and decided to throw my hat into the "ring."</p> <h2>Partial Answers and Observations:</h2> <ol> <li><p><strong>If</strong> the identity element has a countable neighborhood base, <strong>then</strong> trivially by the continuity of the operation of addition, every point has a countable neighborhood base (as continuous maps take a filter base at some point x, and map this onto a filter base of the image point this is a particular example of homogeneity exhibited by topologically equipped algebraic objects.). </p></li> <li><p>Provided, the topology defined on the ring is $T_0$, we have that the topology must also be $T_2$ and completely regular.</p></li> <li><p>Moreover, since there is a countable dense subset $D$ by assumption, we have that given the above two assumptions, there exists a countable collection, of countable covers $U_n$, with the following property: Given any closed set $C$ from this space, and point $p\not\in C$ there exists an $n\in\omega$ such that for some $O\in U_n$, $p\in O$ and $O \cap C = \{\}$. (We get such a collection by ordering the neighborhood base of each point in $D$ by reverse inclusion, and taking fixed 'sections of fixed height' from each to form the open covers)</p></li> </ol> <p>Putting (1), (2), and (3) together produces a topological space which is about as close to a normal Moore space as you can get without actually being one, that is to say, under these assumptions: $R$ is a separable completely regular developable Hausdorff space, (and we haven't even invoked the ACC yet) </p> <h2>The interesting Part:</h2> <p>Ignoring for the moment the previous assumptions, intuitively, the ACC should somehow produce a covering property for this particular space. However, there is an interesting problem when it comes to the definition of subgroup/subring (which is required to get to the notion of ideal needed to apply the ACC): will they be open, closed, neither? Because of this we cannot really apply the ACC, to produce a nice covering property that might have tied everything together (like Lindelöf.) </p> <p><strong>Edit</strong>: While poking around Wiki, I came across something I felt I needed to add</p> <p>However, if you mean that the space is a Noetherian topological space, then we get some gnarlly consequences ( <a href="http://en.wikipedia.org/wiki/Noetherian_topological_space" rel="nofollow">http://en.wikipedia.org/wiki/Noetherian_topological_space</a> ), like the fact that the space is compact! Which is exactly the thread we would want to tie everything together, and produces a normal moore space.</p> <h2>The Reality of the Matter:</h2> <p>The question is ill-posed, in that we do not have enough information to properly deduce a valid and fully general answer. My answer to this question has tried to highlight this point by giving you a case where, you can be about as close as you might ever want to be to something genuinely interesting, and then failing to make it interesting because of the incompatibility of the algebraic assumption with the topological ones. Even if we consider the other possibility we enter one of those strange and beautiful areas in topology where things being to become independent of $ZFC$.</p> <h2>Final Conclusion:</h2> <p>Because of this freedom or lack of information, we are left with an answer of <strong>Most Likely No</strong>. (Weak answer I know) But I can make this claim, because we honestly do not completely understand the notion of hereditary separability (in fact it was only in 2006 that J T Moore was able to produce a ZFC example of an L-space <a href="http://www.math.cornell.edu/~justin/Ftp/Lspace.pdf" rel="nofollow">Article</a>)</p> http://mathoverflow.net/questions/55392/intended-interpretations-of-set-theories/55395#55395 Answer by Michael Blackmon for Intended interpretations of set theories Michael Blackmon 2011-02-14T11:02:01Z 2011-02-14T11:13:33Z <p>To begin, its currently 6:00 am here and I am not entirely alive mentally, so if my answer is missing the point, let me know. In addition, I just thought I might make some additional points or comments not already mentioned in the other answers.</p> <p>This is a common point of contention, and honestly there really isn't a good answer that avoids the issues of playing fast and loose with the consistency of $ZF$. Everyone really just comes up with a meta-mathematical justification which sort of sits the best with them and it really is a matter of philosophy and personality.</p> <p>Firstly, with respect to this comment:</p> <blockquote> <p>This point of view has always made me feel a bit uncomfortable. How can a variable in a first-order language run over the elements of a collection that is not a set? Only recently I realized that one thing is to be a platonist, an another thing is to believe such an odd thing.</p> </blockquote> <p>Formally within the confines of $ZFC$ a variable cannot range over anything that is not a set. However this does not stop us from viewing or speaking about things we cannot formalize within $ZFC$, and Kunen takes liberties with this point of view. The thing that is actually going on when he talks about variables which range over proper classes, is we have just stepped outside of $ZFC$ and are now conversing in the meta-theory.</p> <p>Secondly the comment you make here:</p> <blockquote> <p>To be a bit more concrete, if one is a platonist and the cumulative hierarchy is what one has in mind as the real universe of sets, one can think that the V of one's theory actually refers to a an initial segment of that hierarchy, hence variables run no more over the real V but only over the elements of some $V_{\alpha}$.</p> </blockquote> <p>is a very keen observation, and foreshadows a nice understanding of inaccessible cardinals assumptions. You see by postulating the existence of a strongly inaccessible cardinal we are in fact assuming that we have an initial segment of the cumulative hierarchy which satisfies all of $ZFC$. The only problem with this is that in doing so we have stepped up the consistency strength of system we are working in.</p> <p>Thirdly, your next comment</p> <blockquote> <p>There's a parallel to these ideas. For example, when we want to prove consistency with ZFC of a given sentence, we do not directly look for a model of ZFC where that sentence is true, but instead we take advantage of knowing that every finite fragment of ZFC is consistent and that every proof involves only finitely many axioms.</p> </blockquote> <p>is quite dead on. Because applying Levy reflection, coupled with the downward Löwenheim–Skolem theorem we can produce a countable model of enough of $ZFC$ for whatever argument we care to be trying to formulate. This view provides a lot of comfort when trying to construct forcing arguments. However, if this view is not comforting, there is a differing view on the matter, and it makes an appearance in the way of Boolean-valued models (which Kunen's text is kinda lite on).</p> <p>EDIT: fixed some dumb typos.</p> http://mathoverflow.net/questions/52897/singular-cardinals-and-a-strange-question Singular Cardinals, and A Strange Question. Michael Blackmon 2011-01-23T02:42:31Z 2011-02-07T04:03:55Z <p>Let $\mu$ be any infinite cardinal, and define a collection $N\subset[\mu]^\mu$ to be, maximal almost disjoint (MAD) over $\mu$, iff </p> <ol> <li>$\forall\{A,B\}\in[N]^2$ $( A\cap B \in [\mu]^{&lt;\mu})$</li> <li>$\forall X\in[\mu]^\mu \exists A\in N$ $( X \cap A \in [\mu]^\mu)$</li> </ol> <p>My questions are as follows: when $\mu$ is <strong>singular</strong></p> <ol> <li>What is known about MAD families (or any other combinatorial structures, like, towers, SFIP families without pseudo-intersection, etc) over $\mu$?</li> <li>Are such families degenerate in the sense that an infinite family can have cardinality below $\mu$? </li> <li>Is there any connection between such constructs on $\mu$ and the corresponding constructs on $cf(\mu)$?</li> </ol> <p>The main point I really want to know is this: Is it possible to add new subsets to an arbitrary singular cardinal without adding new subsets the the cardinals below it? </p> <p>Side Request: I've been told there are forcing constructions which will add an order type $\omega$ cofinal sequence to a cardinal with cofinality $\omega$, can anyone point me in the correct direction with a book or article?</p> http://mathoverflow.net/questions/54399/if-a-result-is-apparently-provable-with-ac-is-actually-independent-of-zf/54455#54455 Answer by Michael Blackmon for If a result is apparently provable with AC, is actually independent of ZF? Michael Blackmon 2011-02-05T19:33:29Z 2011-02-06T19:16:07Z <p>There are many such proofs which use $AC$, but which are not even close to being independent of $ZF$. The general reason for this is how $AC$ is used. Normally the non-essential use of $AC$ appears when you have a very targeted application in mind, with additional structure in the background.</p> <p>For example: Given an arbitrary collection of non-empty sets $\{X_\alpha: \alpha \in Y\}$ asserting that the product $Z =\Pi_{\alpha \in Y} X_{\alpha}$ is non-empty requires $AC$ when you have no structure imposed on the $X_\alpha$ and $Y$. However, when we add the assertion that each $X_\alpha$ is a ring, with additive identity $0_\alpha\in X_\alpha$, we then know that $Z$ is always non-empty without $AC$. The reason for this is because we now can define a function which witnesses that $Z$ is non-empty, in fact the function $\varphi:Y \rightarrow \bigcup X_\alpha$, given by $\varphi(\alpha) = 0_\alpha$ is such a witness, because $\varphi \in Z$.</p> <p>That having been said, as for a specific example of a theorem for which everyone thought relied on $AC$ but was proven to hold in $ZF$, I cannot think of one off-hand that has not already been mentioned. But I think an example of what you are looking for might be contained in a question by Andres Caicedo, <a href="http://mathoverflow.net/questions/40507/distinct-well-orderings-of-the-same-set/" rel="nofollow">http://mathoverflow.net/questions/40507/distinct-well-orderings-of-the-same-set/</a> and in his insightful answer to his own question.</p> http://mathoverflow.net/questions/54232/a-book-you-would-like-to-write/54256#54256 Answer by Michael Blackmon for A Book You Would Like to Write Michael Blackmon 2011-02-03T22:53:10Z 2011-02-03T23:11:11Z <p><strong>Book Title:</strong> <em>An Introduction to Forcing (for people who don't care about foundations.)</em></p> <p><strong>Synopsis</strong>: Forcing is one of the most amazing techniques in use today, and it offers amazing insight into how objects in mathematics can be constructed. The aim of this book would be to focus on the tools and methods of Forcing, and provide examples of constructions which highlight the intrinsic beauty that can be found hiding under the surface of a forcing argument. Moreover, it would highlight the practical applications of, and sense of naturalness the "Forcing Perspective" brings to inductive mathematical constructions (which might be outside the domain of set-theoretic interest.)</p> <p><strong>Reason For Wanting to Write It:</strong> When I first learned about Forcing, the first thing that struck me was <em>"Why the hell has no one ever told me about this? What the hell!? This is AWESOME!"</em> That sense of awe has stayed with me throughout my very short "career." So the book would be a way for me to share this view with other mathematicians who don't really care all that much about "set theory", "category theory", or "foundations" (just like I did before I learned about independence proofs, etc.) Moreover, the aim would not be to convert them to some relativist view of mathematics, but to just show them how directly linking the logical structure of an object with its construction can open new doors, and add much needed perspective to any field. </p> <p><strong>When Would It Get Written:</strong> Honestly, not now, and not in the near future, maybe 10/20 years. The reason for this is, I just don't know enough yet, I'm still a student. That being said, I must admit, I am most likely not the first person anyone would pick to write such a book. However, if I was ever presented with the opportunity I would take it in a heartbeat. To me the importance of the ideas and perspective for mathematics as a whole out weigh the possible huge list of errors and corrections that would follow such a book (if written by me that is).</p> <p>PS: if there are any spelling or grammar errors, feel free to fix them.</p> http://mathoverflow.net/questions/53711/cn-and-forcing-reading-a-recent-paper-by-kunen $C^n$ And Forcing: Reading a Recent Paper By Kunen Michael Blackmon 2011-01-29T13:17:21Z 2011-01-31T06:11:19Z <p>While reading a recent paper by Kunen <a href="http://arxiv.org/abs/0912.3733" rel="nofollow">arxiv.org/abs/0912.3733</a>, which deals with PFA and the existence of certain differentiable functions, (defined on all of $\mathbb{R}$) which map certain $\aleph_1$-dense subsets of $\mathbb{R}$ onto other $\aleph_1$-dense subsets of $\mathbb{R}$. The technical details aside, he was able to show:</p> <blockquote> <p><strong>Theorem 1.6</strong> Assume PFA, and let $D,E,\subset \mathbb{R}$ be $\aleph_1$-dense. Then there exist exists an order preserving bijection $f:\mathbb{R} \rightarrow \mathbb{R}$, and $D^\ast \subseteq D$ such that $D^\ast$ is $\aleph_1$-dense, $f(D^\ast)=E$, and </p> <ol> <li>For all $x \in \mathbb{R}$, $f'(x)$ exists and $0\leq f'(x)\leq 2$</li> <li>$f'(d) = 0$ for all $d\in D^\ast$</li> </ol> </blockquote> <p>It occurred to me that with a few modifications his method/forcing notion might be used to add other differentiable, or Lipschitz functions to some ground model. It follows that these new functions would in turn produce new $C^1$ functions, and so on. And in the end <strong>could</strong> result in new systems of differential equations, with absolutely strange behavior.</p> <p>So my questions are the following:</p> <ul> <li>What is known about "messing with" the class of $C^n$ functions, via forcing?</li> <li>Are there other examples of more exotic forcing notions which add smoother functions?</li> </ul> <p>Edit: took out weakly bit..</p> http://mathoverflow.net/questions/53696/an-ordinal-invariant-for-spaces-based-on-a-hierarchy-of-closed-sets-from-z-sets/53704#53704 Answer by Michael Blackmon for An ordinal invariant for spaces based on a hierarchy of closed sets from z-sets Michael Blackmon 2011-01-29T09:47:39Z 2011-01-30T00:14:53Z <p>EDIT: This is for the normal case....</p> <ol> <li><p>$V \in H(1) \cup H(0)$, if and only if $V$ is a zero set of $X$. (this is by definition) So in particular there exists some $f_V \in C(X,[0,1])$ such that $f_V(x) = 0 \iff x \in V$.</p></li> <li><p>If $U \in H(2)$, then there exists some $V \in H(1) \cup H(0)$, and $g_U\in C(V,[0,1])$ such that $g_U(x) = 0 \iff x \in U$</p></li> <li><p>Because $U$ a zero set of $V$, it will be closed, and so because $X$ is normal, we may apply Tietzes' Extension Theorem to produce a continuous map $h_{UV}:X\rightarrow[0,1]$ which extends $g_U$.</p></li> <li><p>Both the functions $h_{UV}$ and $f_V$ are defined on all of $X$, so we may add them to produce the new function $F:X\rightarrow[0,2]$ given by $F(x)=h_{UV}(x) + f_{V}(x)$.</p></li> <li><p>It follows that $U$ is the zero set of $F$, and we have that $U$ is a zero set of $X$. </p></li> </ol> <p>By (1), (2), (3), (4) and (5), it follows that $U \in H(0) \cup H(1)$ and so $H(2) = H(1) \cup H(0)$. So noting that $X$ is always a zero set of $X$, we have that $H(2) = H(1)$.</p> <p>Therefore, $\nu(X) = 1$</p> <p>PS: If you are looking for a hard problem in this form that deals with topology, may I suggest Alan Dows' "Sequential order" chapter from "Open Problems in Topology II"</p> <p>Edit: attempt at general case removed due to error.</p> http://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts/53709#53709 Answer by Michael Blackmon for Awfully sophisticated proof for simple facts Michael Blackmon 2011-01-29T11:55:08Z 2011-01-29T12:16:42Z <p>Because for some <em>reason</em> no one has mentioned it. </p> <p><strong>Russell's proof that 1+1=2.</strong></p> <p><a href="http://quod.lib.umich.edu/cgi/t/text/pageviewer-idx?c=umhistmath;cc=umhistmath;rgn=full%20text;idno=AAT3201.0001.001;didno=AAT3201.0001.001;view=pdf;seq=00000412" rel="nofollow">http://quod.lib.umich.edu/cgi/t/text/pageviewer-idx?c=umhistmath;cc=umhistmath;rgn=full%20text;idno=AAT3201.0001.001;didno=AAT3201.0001.001;view=pdf;seq=00000412</a></p> http://mathoverflow.net/questions/53673/why-is-it-ok-to-rely-on-the-fundamental-theorem-of-arithmetic-when-using-godel-nu/53693#53693 Answer by Michael Blackmon for Why is it OK to rely on the Fundamental Theorem of Arithmetic when using Gödel numbering? Michael Blackmon 2011-01-29T06:15:36Z 2011-01-29T06:22:41Z <p>It's a trick of encoding. You encode the formulas of the theory into a fixed collection of integers which have a certain form. </p> <p>Now, you have defined somewhere in the background a natural notion of multiplication, addition, and equality of integers. So you are free to ask questions about whether or not two numbers are equal, and whether or not for some number $x$, there is a formula in the language of PA, for which $x$ is the encoding of, and this allows you to go backwards, in the <em>metatheory.</em> (that is take a number and see if it encodes a sentence)</p> <p>That being said, you are never starting with a number and going backwards in the formal theory (that is to say turning it into the sentence it encodes, as this would require FTA), you are always going forward, taking the Gödel encoding of a given sentence.</p> <p>It's this careful dance and intermingling of the syntax and semantics that makes the proof of this theorem so much fun to go over.</p> <p>Hope this clears it up a little bit. I know it probably doesn't completely answer your question.</p> http://mathoverflow.net/questions/26416/what-is-your-favorite-proof-of-tychonoffs-theorem/53584#53584 Answer by Michael Blackmon for What is your favorite proof of Tychonoff's Theorem? Michael Blackmon 2011-01-28T09:09:42Z 2011-01-28T09:09:42Z <p>Personally, I've always enjoyed the proof given in Topology, by Hocking and Young. It's essentially the basic ultrafilter proof, but its got a nice feel to it. I guess I'm biased because this was the first real Topology book I was ever able to get my hands on.</p> http://mathoverflow.net/questions/5499/which-mathematicians-have-influenced-you-the-most/53460#53460 Answer by Michael Blackmon for Which mathematicians have influenced you the most? Michael Blackmon 2011-01-27T07:04:48Z 2011-01-27T12:59:55Z <ul> <li>Paul <strong>Cohen</strong> &amp; Kurt <strong>Gödel</strong> <ul> <li>They gave us the tools to construct models of set theory.</li> </ul></li> <li>Kenneth Kunen <ul> <li>His book "Set theory: An Introduction To Independence Results" was the book that got me interested in the field I would later call my home.</li> </ul></li> <li>Saharon <strong>Shelah</strong> <ul> <li>His work on forcing, and singular cardinals keep me asking questions, and open up the possibility for questions I didn't even know could be asked.</li> </ul></li> </ul> http://mathoverflow.net/questions/52032/examples-of-non-metrizable-spaces/53319#53319 Answer by Michael Blackmon for Examples of non-metrizable spaces Michael Blackmon 2011-01-26T05:16:18Z 2011-01-27T07:27:06Z <p>Because nobody has mentioned it, and in my mind, this is one of the biggest surprises set-theoretic topology has ever given mathematics, here it is:</p> <p>The following is independent of ZFC: <strong>Every normal Moore space is metrizable.</strong></p> <h2>Background</h2> <p>Technically speaking, a Moore space is a developable regular Hausdorff space. To see that Moore spaces can occur in nature, and kind of get a feel for what they are consider my answer to this question <a href="http://mathoverflow.net/questions/7103/topological-rings/53145#53145" rel="nofollow">http://mathoverflow.net/questions/7103/topological-rings/53145#53145</a> .</p> <h2>Some Results</h2> <ul> <li>Assuming $CH$, there is a normal Moore space which fails to be metrizable. </li> </ul> <p>Reference: William G. Fleissner's <a href="http://www.ncbi.nlm.nih.gov/pmc/articles/PMC345971/" rel="nofollow">"Normal nonmetrizable Moore space from continuum hypothesis or nonexistence of inner models with measurable cardinals"</a></p> <ul> <li>If every normal Moore space is metrizable, then there exists an inner model with a measurable cardinal</li> </ul> <p>Reference: William G. Fleissner's <a href="http://www.jstor.org/pss/1999211" rel="nofollow">"If All Normal Moore Spaces Are Metrizable, Then There Is An Inner Model With A Measurable Cardinal"</a></p> http://mathoverflow.net/questions/49151/most-intricate-and-most-beautiful-structures-in-mathematics/53313#53313 Answer by Michael Blackmon for Most intricate and most beautiful structures in mathematics Michael Blackmon 2011-01-26T02:39:05Z 2011-01-26T02:39:05Z <p>Shelah's Body of Work. Considering that <a href="http://shelah.logic.at/shelah_a.dvi" rel="nofollow">this</a> list of references is over 100 pages long, I think this a contender.</p> http://mathoverflow.net/questions/40507/distinct-well-orderings-of-the-same-set/53064#53064 Answer by Michael Blackmon for Distinct well-orderings of the same set Michael Blackmon 2011-01-24T14:37:14Z 2011-01-24T14:54:33Z <p><strong>I am hesitant to attempt this, so If this makes no sense, or is not in keeping with what you are looking for I'm sorry.</strong></p> <p>Firstly, let $R_1, R_2 \subset \kappa\times\kappa$, be the two well-orderings of $\kappa$ we are considering. Then, as $\kappa \times \kappa$ has a well-ordering $\lt_O$ which has order type with $\kappa$ via some isomorphism $P:\kappa \times \kappa \rightarrow \kappa$, with for each $(\alpha,\beta)\in \kappa \times \kappa$:</p> <p>$\gamma = max(\alpha,\beta) \implies P((\alpha,\beta)) = P(\alpha,\beta) \in P''((\gamma+1)\times(\gamma+1)).$</p> <p>(This is a modification of the global ordering from Set Theory and The Continuum Hyp. by R. M. Smullyan and M. Fitting, p. 119).</p> <p>This ordering is defined as follows: $(\alpha,\beta)\lt_O(\gamma,\delta)$ iff one of the following holds</p> <ul> <li>$max(\alpha,\beta) \lt max(\gamma,\delta)$, or</li> <li>$max(\alpha,\beta) = max(\gamma,\delta)$, and $\alpha \lt \gamma$, or</li> <li>$max(\alpha,\beta) = max(\gamma,\delta)$, and $\beta \lt \delta$</li> </ul> <p>As such, we may form the following construction: Let $r^k_0=min_{\lt_O} (R_k)$ and for $\alpha \lt \kappa$ define $r^k_\alpha = min_{\lt_O} (R_k \backslash \{r^k_\gamma :\gamma \lt \alpha\} )$ for $k=1,2.$</p> <p>Next following along with the answer from, Alessandro Sisto, we may define the sequence $u_\alpha$ </p> <p>$u_\alpha = min\{ \beta \in P''(\alpha+1)\times(\alpha+1): \forall \gamma \lt \alpha ( P^{-1}(u_{\gamma}) \lt_O r^1_{\beta} \text{ and } P^{-1}(u_\gamma) \lt_O r^2_\beta) \}.$</p> <p>Then $u_\alpha$ will be defined for every $\alpha \lt \kappa$, and so preforming the same bit that Alessandro did, with $y_\alpha = min \{ \beta \in \kappa: \exists \gamma \lt \alpha( (\eta,\beta) = P^{-1}(y_\gamma)) \}$ should produce the result.</p> http://mathoverflow.net/questions/52901/ultrapowers-and-models-of-set-theory Ultrapowers, and Models of Set theory. Michael Blackmon 2011-01-23T03:21:30Z 2011-01-23T06:38:16Z <p>Let $V$ be the set-theoretic universe, and suppose that $U$ is some ultrafilter over $\kappa\gt\omega.$ Then, we can go through the motions and produce the ultrapower $M = V^{\kappa}/U$. </p> <p>Now, the existence of an $\omega_1$-complete ultrafilter (and the existence of the transitive collapse of $M$) is subject to (as I currently understand it) the existence of a measurable cardinal. So my questions are:</p> <ol> <li>in the absence of a strong large cardinal hypothesis (like the existence of a measurable cardinal) what can we say about the ultrapower M? </li> <li>Is it possible to recover or produce some kind of embedding (which will not be a full elementary embedding) of $V$ into $M$, which preserves enough structure to work within $M$?</li> </ol> http://mathoverflow.net/questions/52902/proving-theorems-by-using-functions-with-fixed-points/52903#52903 Answer by Michael Blackmon for Proving theorems by using functions with fixed points. Michael Blackmon 2011-01-23T03:52:22Z 2011-01-23T03:52:22Z <p>The Fixed point phenomenon has a lot of very useful applications. You mentioned the Contraction mapping theorem, well that particular theorem pops up all over the place for example: in the proofs of </p> <ol> <li>The Stable/Unstable Manifold Theorem, for systems of differential equations.</li> <li>The Hartman-Grobman Theorem, which relates the local behavior of a system of differential equations to the linearized system in a neighborhood of a hyperbolic singularity.</li> </ol> <p>The Inverse Function Theorem also crops up everywhere, (but mainly in non-constructive existence proofs): In particular, in the proofs of</p> <ol> <li>The Existence of the Poincare First Return Map, for a periodic orbit of a system of differential equations</li> <li>The Flowbox Theorem</li> <li>Finding Singular Points for surfaces/curves, (these are points where the Jacobian has determinate zero)</li> </ol> <p>These are all I can think of at the moment.</p> http://mathoverflow.net/questions/42873/strange-question-about-hechler Strange question about Hechler Michael Blackmon 2010-10-20T04:35:14Z 2010-10-20T18:11:28Z <p>Recently during a lecture, my professor mentioned that forcing over any poset which is countable, separative, and atomless, is essentially the same as forcing over the Cohen poset, that is to say results in adding a Cohen real. </p> <p>My question is: Are there any other similar characterizations of "commonly used" forcing posets? Specifically, is there one for the Hechler poset?</p> <p>The Hechler Poset/forcing notion $(H,\le)$ is given by setting $H=\omega^{\lt\omega} \times \omega^{\omega}$, and defining the relation $(t,v) \le (s,u)$ iff $( t \supset s \wedge (\forall n\in\omega) (u(n) \le v(n)) \wedge (\forall m \in dom(t \backslash s))(t(m) \gt u(m))$. When forcing with this poset, you end up adding an unbounded real to the ground model.</p> <p>I understand that you cannot produce a model in which $\mathfrak{b}=\omega_2$ using product forcing, and that you need iterated forcing to do so. Moreover, the iterated forcing construction I've seen that produces $\mathfrak{b}=\omega_2$ in the forcing extension, used the finite support iteration of $\omega_2$ many copies of the Hechler poset. Is this evidence for the lack of such a characterization?</p> <p>(I apologize in advance if this is an ill-stated question, I will change it accordingly if it is.)</p> http://mathoverflow.net/questions/42590/independence-and-category-theory Independence and Category Theory Michael Blackmon 2010-10-18T07:14:19Z 2010-10-19T05:25:19Z <p>I'm not very experienced with respect to Category Theory. So if this question makes no sense I'm sorry. At any rate here is my question: If the existence or non-existence of specific sets can be independent of set theory, then how can it be that the category Set is complete under small limits?</p> <p>For example, suppose you have a small category A, with objects that are linearly ordered spaces X such that: X is without smallest or largest element, X has CCC, X is complete, and X is dense in itself. And as morphisms for A, you take order preserving bijections. Now, let F be the forgetful functor from A to the underlying set.</p> <p>How exactly can you define the limit over F inside of Set?</p> <p>Another example, would be considering some set sized collection of Whitehead groups, call it X. Now, consider X as a category equipped with homomorphisms as morphisms. And let G be the forgetful functor from X into Set.</p> <p>How exactly can one define the limit over G inside Set?</p> <p>Or even better, suppose that G is the identity functor from X into Grps. Then, what happens?</p> <p>Am I correct in saying that the answer depends drastically on the set theoretic universe you pick? If so, how is this not a problem with category theory?</p> http://mathoverflow.net/questions/122010/subalgebra-of-a-simple-forcing Comment by Michael Blackmon Michael Blackmon 2013-04-02T01:55:24Z 2013-04-02T01:55:24Z (Where the completion of $Add(\omega,\alpha)\ast Add(\omega_1,1)$ is taken to be corresponding regular open algebra) http://mathoverflow.net/questions/122010/subalgebra-of-a-simple-forcing Comment by Michael Blackmon Michael Blackmon 2013-04-02T01:53:56Z 2013-04-02T01:53:56Z I strongly suspect you mean that $\Vert \check{\beta} \in \dot{X} \Vert = \{ (p,\dot{q}) \in Add(\omega,\alpha)\ast Add(\omega_1,1): p \Vdash (\beta, 1) \in \dot{q} \}$ http://mathoverflow.net/questions/118407/forcing-over-the-poset-of-nonempty-open-subsets-of-a-nice-topological-space/118409#118409 Comment by Michael Blackmon Michael Blackmon 2013-01-09T01:47:50Z 2013-01-09T01:47:50Z @Adam Epstein, yes and by infinite I mean infinite dimensional. http://mathoverflow.net/questions/118407/forcing-over-the-poset-of-nonempty-open-subsets-of-a-nice-topological-space/118409#118409 Comment by Michael Blackmon Michael Blackmon 2013-01-09T01:16:08Z 2013-01-09T01:16:08Z @Adam Epstein, a more fruitful exercise would be to instead focus on infinite CW complexes and produce new ones generically, then ask just how much you can monkey around with the constructions of algebraic topology. http://mathoverflow.net/questions/114820/type-i-subspaces-of-the-stone-cech-compactification-of-omega Comment by Michael Blackmon Michael Blackmon 2012-11-28T23:46:50Z 2012-11-28T23:46:50Z K. Kunen proved that there are always weak P-points in $\omega^{\ast}$, (Note: a weak P-point is a point $p$, which is not in the closure of any-countable subset of $\omega^{\ast}\backslash \{p\}$ . This should be enough to make a modified version of your argument work, where you use nested countable subsets of $\omega^{\ast}$ whose closures avoid the weak P-point, to build the sequence $X_\alpha$. http://mathoverflow.net/questions/112466/can-assumptions-about-forcing-produce-mice Comment by Michael Blackmon Michael Blackmon 2012-11-19T06:19:33Z 2012-11-19T06:19:33Z Mariano, sorry: Just learned life is a two player game, and the only sane ref is logic. http://mathoverflow.net/questions/112466/can-assumptions-about-forcing-produce-mice Comment by Michael Blackmon Michael Blackmon 2012-11-19T05:35:53Z 2012-11-19T05:35:53Z Its very complicated sorry. http://mathoverflow.net/questions/112466/can-assumptions-about-forcing-produce-mice/112510#112510 Comment by Michael Blackmon Michael Blackmon 2012-11-15T22:56:33Z 2012-11-15T22:56:33Z Andreas, Thank you so much, I owe you a ladder system on \omega_1. http://mathoverflow.net/questions/112466/can-assumptions-about-forcing-produce-mice/112489#112489 Comment by Michael Blackmon Michael Blackmon 2012-11-15T19:38:04Z 2012-11-15T19:38:04Z Oh no, don't get me wrong that is exactly where my intuition about forcing comes from and is why I was a bit worried, since proofs always carry more weight than intuitions. http://mathoverflow.net/questions/112466/can-assumptions-about-forcing-produce-mice/112489#112489 Comment by Michael Blackmon Michael Blackmon 2012-11-15T18:07:22Z 2012-11-15T18:07:22Z (and makes me finally feel like I can sanely use forcing to produce consistency results again.) http://mathoverflow.net/questions/112466/can-assumptions-about-forcing-produce-mice/112489#112489 Comment by Michael Blackmon Michael Blackmon 2012-11-15T18:01:42Z 2012-11-15T18:01:42Z Joel, Thank you very much for the counter-example; this question has been bothering me for a couple of weeks, and for the life of me I couldn't figure out how to construct a forcing with these properties without first assuming either Con(ZFC) or \neg Con(ZFC); but as you seem to hint at, this is not at a problem because the extension of the forcing language to include the constant $\check{\lambda}$ for a fixed regular cardinal $\lambda$ is necessarily a more expressive language than that without constants. In particular forcing cheats way harder than I'd ever expected. http://mathoverflow.net/questions/112466/can-assumptions-about-forcing-produce-mice Comment by Michael Blackmon Michael Blackmon 2012-11-15T11:50:01Z 2012-11-15T11:50:01Z Well, I didn't say $1 \forces_\mathbb{P} \lambda$ is measurable. I took a single consequence of a measurable cardinals existence: namely changing the cofinality of a regular cardinal without collapsing cardinals. and asked if that was the same as asserting a measurable exists (For what its worth singular cardinals, can't be measurable (club filter not being sufficiently closed and all.) so in the extension that cardinal is actually no longer measurable. Which is a subtle point you don't seem to be getting. http://mathoverflow.net/questions/112466/can-assumptions-about-forcing-produce-mice Comment by Michael Blackmon Michael Blackmon 2012-11-15T11:26:24Z 2012-11-15T11:26:24Z Key point here is the word Witness. In particular, a model does not satisfy a particular existential statement without first being able to produce a witness certifying that statement holds. http://mathoverflow.net/questions/112466/can-assumptions-about-forcing-produce-mice Comment by Michael Blackmon Michael Blackmon 2012-11-15T11:21:34Z 2012-11-15T11:21:34Z I just want to know if a statement asserting the existence of a regular cardinal and particular partial order, has the same strength as the statement there exists a regular cardinal with a normal measure. http://mathoverflow.net/questions/112466/can-assumptions-about-forcing-produce-mice Comment by Michael Blackmon Michael Blackmon 2012-11-15T11:19:28Z 2012-11-15T11:19:28Z $\aleph_{\omega_1}$ is not regular.