User arthur fischer - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T04:46:14Z http://mathoverflow.net/feeds/user/13653 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92023/two-versions-of-absolutely-ccc Two versions of "absolutely ccc" Arthur Fischer 2012-03-23T17:04:33Z 2013-05-03T11:22:00Z <p>I have recently been slogging my way through Shelah's "<em>Large continuum, oracles</em>". Essentially from the start there has been a question needling me which I cannot seem to answer.</p> <ul> <li>In the paper, Shelah says that a forcing notion $\mathcal{P}$ is <em>absolutely ccc</em> if it remains ccc after forcing with any ccc notion.</li> <li>Elsewhere, I have seen it defined that a forcing notion $\mathcal{P}$ is <em>absolutely ccc</em> if it remains ccc after any forcing. (This would be <em>indestructibly ccc</em> from BartoszyĆski-Judah.)</li> </ul> <p>Any forcing having the Knaster property is absolutely ccc (in the strong sense), and MA$_{\aleph_1}$ implies that all ccc forcings have the Knaster property. Thus, it is consistent that the two are equivalent.</p> <p>Do these two versions of absolute ccc-ness provably coincide?</p> http://mathoverflow.net/questions/122382/how-to-see-such-space-is-lindelof/122383#122383 Answer by Arthur Fischer for How to see such space is Lindelof? Arthur Fischer 2013-02-20T08:31:43Z 2013-02-20T08:31:43Z <p>Note that the open subsets of (what I will denote by) $\mathbb{R}_B$ are of the form $U \cup A$ where $U \subseteq \mathbb{R}$ is open in the usual topology, and $A \subseteq B$ is arbitrary. </p> <p>Suppose that $\{ U_i \cup A_i : i \in I \}$ is an open cover of $\mathbb{R}_B$. Note that there is a countable $I_0 \subseteq I$ such that $\bigcup_{i \in I_0} U_i = \bigcup_{i \in I} U_i$. Next note that $\mathbb{R} \setminus \bigcup_{i \in I} U_i \subseteq B$ is closed (in $\mathbb{R}$) and is therefore countable, so there is a countable $I_1 \subseteq I$ such that $\mathbb{R} \setminus \bigcup_{i \in I} U_i \subseteq \bigcup_{i \in I_1} A_i$.</p> http://mathoverflow.net/questions/109205/existence-of-weakly-compact-cardinals/109212#109212 Answer by Arthur Fischer for Existence of weakly compact cardinals Arthur Fischer 2012-10-09T09:19:44Z 2012-10-09T09:19:44Z <p>As weakly compact cardinals are in particular <a href="http://en.wikipedia.org/wiki/Inaccessible_cardinal" rel="nofollow">(strongly) inaccessible</a>, it follows from Gödel's Second Incompleteness Theorem that ZFC cannot prove the implication "Con(ZFC) implies Con (ZFC + $\exists$weakly-compact )." (Unless, of course, if ZFC is itself inconsistent, at which point this is all a lot of bunk).</p> <p>(The linked Wikipedia article outlines the basic reasoning.)</p> <p>Any argument for the consistency of "ZFC + $\exists$weakly-inaccessible" must therefore transcend ZFC. (But, as mentioned by Asaf, no contradictions have been found thus far been under the assumption of the existence of weakly-compact cardinals.)</p> http://mathoverflow.net/questions/107163/a-function-that-is-defined-everywhere-but-has-unknown-values/107176#107176 Answer by Arthur Fischer for A function that is defined everywhere but has unknown values Arthur Fischer 2012-09-14T13:03:48Z 2012-09-14T13:13:24Z <p>Kind of a cheat, but define $f : \mathbb{N} \to \mathbb{N}$ so that $f(n)$ is the initial position (to the right of the decimal point) of the first occurance of $n$ consecutive $5$s in the the decimal expansion of $\pi$, and $0$ if such does not exist.</p> <p>So $f(1) = 4$, $f(2) = 130$, $f(3) = 177$, $f(4) = 24,466$, etc. I don't think I'm saying too much by claiming that as it is unknown whether $\pi$ is normal, we do not know if $f(n)$ is non-zero for all $n$.</p> <p><strong>EDIT:</strong> I honestly didn't see the almost identical answer in the comments above before posting this. It has thus been made CW.</p> http://mathoverflow.net/questions/105117/countable-topological-spaces-of-uncountable-weight/105142#105142 Answer by Arthur Fischer for countable topological spaces of uncountable weight Arthur Fischer 2012-08-21T05:17:18Z 2012-08-21T05:17:18Z <p>Another classical example is the Arens-Fort Space.</p> <p>Let $X = \omega \times \omega$. For each $A \subseteq X$ and $n \in \omega$ we let $A_n = \{ m \in \omega : (n,m) \in A \}$ denote the $n$th section of $A$.</p> <p>Topologise $X$ by taking each point of $X \setminus \{ (0,0) \}$ to be isolated, and let $U \subseteq X$ be a neighbourhood of $(0,0)$ iff it contains $(0,0)$ and all but finitely many sections of $U$ are co-finite.</p> <p>Clearly this space is T$_1$. If $F, E \subseteq X$ are disjoint closed sets, then one, say $E$, does not contain $(0,0)$. So $E$ is clopen, and $X \setminus E$ is an open set including $F$ which is disjoint from $E$. Thus $X$ is normal.</p> <p>To show that $X$ is not second countable, it suffices to show that there is no countable base at $(0,0)$. If $\{ U^{(i)} \}_{i \in \omega}$ is any family of open neighbourhoods of $(0,0)$, we inductively define a sequence of pairs of natural numbers $\{ (n_i,m_i) \}_{i \in \omega}$ so that:</p> <ul> <li>$n_i > n_{i-1}$ is such that $U^{(i)}_{n_i}$ is non-empty (say $n_{-1} = 0$); and</li> <li>$m_i \in U^{(i)}_{n_i}$.</li> </ul> <p>Then $V = X \setminus \{ ( n_i , m_i ) : i \in \omega \}$ is an open neighbourhood of $(0,0)$, and by construction $U^{(i)} \not\subseteq V$ for all $i$.</p> http://mathoverflow.net/questions/124494/bijective-equivalent-collections-of-proper-classes-in-set-theory Comment by Arthur Fischer Arthur Fischer 2013-03-14T10:43:31Z 2013-03-14T10:43:31Z What sort of axiomatization are you thinking of for NBG/MK? I think the most common axiomatizations include (or imply) Limitation of Size, which says that a class is proper iff it admits an injection from V, and from which you then get Global Choice. http://mathoverflow.net/questions/122382/how-to-see-such-space-is-lindelof/122383#122383 Comment by Arthur Fischer Arthur Fischer 2013-02-20T09:57:50Z 2013-02-20T09:57:50Z @John: Do you mean my characterisation of the open subsets of $\mathbb{R}_B$? (Which follows from the fact that the topology generated by the usual open subsets of $\mathbb{R}$ and the singletons from $B$.) Or that there is a countable $I_0$? (Which follows from the fact that $\mathbb{R}$ is second-countable, and thus hereditary Lindel&#246;f.) http://mathoverflow.net/questions/104731/instances-where-an-existence-result-precedes-the-constructive-version/105053#105053 Comment by Arthur Fischer Arthur Fischer 2012-08-20T17:24:11Z 2012-08-20T17:24:11Z Didn't Hilbert himself produce a constructive proof a few years after his non-constructive one? http://mathoverflow.net/questions/92023/two-versions-of-absolutely-ccc Comment by Arthur Fischer Arthur Fischer 2012-03-24T08:01:49Z 2012-03-24T08:01:49Z @saf: See <i>e.g.</i>, Jech (3rd ed.), Theorem 16.21, p.277. The proof actually gives the slightly stronger result that MA$_{\aleph_1}$implies that all ccc posets have precalibre $\aleph_1$. http://mathoverflow.net/questions/92023/two-versions-of-absolutely-ccc/92027#92027 Comment by Arthur Fischer Arthur Fischer 2012-03-24T06:36:15Z 2012-03-24T06:36:15Z To further Joel's point (or confusion) I believe that the usual construction of a Hausdorff gap yields an indestructible gap. It thus seems that this answer would contradict what I've stated in the question about the two formulations of absolute ccc-ness being consistently equivalent.