User goldstern - MathOverflow

Answer by Goldstern for Is it possible to reconstruct an order type from its initial segments?

2013-05-06T21:59:41Z

No. Take $\omega_1$, with each element replaced by a copy of $\mathbb Q$. Then $S$ will contain a single order type. (The rest is left as an exercise.)

Answer by Goldstern for A question about "paradoxical" sentences in the language of ZF set theory.

2013-05-05T18:34:57Z

You are looking for a definable class $C$ such that (ZFC-provably) neither $C$ nor its complement can be a set. There are lots of those. Note that a class is not a set iff it contains elements of arbitrarily high rank. If you define a class at random, I would think the chances are at least 99% that it has this property.

Some examples.

The class of all sets containing your favorite set $s_0$. (As an element. Or, as a subset - unless $s_0=\emptyset$.)
The class of all groups. (Or, your favorite class of structures, unless it happens to be the class of all sets.)
The class of all finite sets. The class of all sets of size $\kappa$.
The class of all sets of even rank.

No, this is too easy, I must have misunderstood the question...

Answer by Goldstern for What is the best general triangle?

2013-04-25T15:40:22Z

The book by "Humor in der Mathematik" by Friedrich Wille (from the 1970s or 1980s) contains the tongue-in-cheek theorem "Up to similarity, there is a unique general triangle". (Google book search for "Friedrich Wille" "Allgemeines Dreieck")

"General" is defined as "all angles must differ from each other, and from 90 degress, by at least 15 degrees".

Given some further axioms (like: diagonals of acute angles have to be visually distinct from line of symmetry) it is also shown that there is a unique general quadrilateral, and that this quadrilateral is "teeming" (another technical term) with general triangles.

Answer by Goldstern for A Fraïssé class without the strong amalgamation property.

2013-04-11T23:08:08Z

I am not sure what this is good for (and if I understand your terminology), but here is a trivial example: Take a language with one unary predicate $P$, and let $F$ be the class of all finite structures in which there is at most one element $e$ such that $P(e)$ holds.

If I am not mistaken, the family of (finite) distributive lattices is a less trivial example.

Answer by Goldstern for Name convention for the composition of the preimage of a function and the function itself

2013-04-07T19:54:51Z

I do not know any name for it. However, note that the concept you are definining does not really depend on the function $f$, only on the equivalence relation induced by $f$ (which is sometimes called the kernel of $f$, unless you work with groups).

A set $S$ is called "saturated" with respect to an equivalence relation $\theta$ iff $S$ is a union of equivalence classes, or equivalently, if $S=f^{-1}(f(S))$ (where $f$ is some map inducing $\theta$).

Hence "saturation" (as Benjamin Dickman suggested in a comment) would be a natural choice.

But if the audience members are not mathematicians, "saturation" might mean something entirely different to them. I would suggest "$f$-neighborhood"; this term can be easily visualized, I think.

Answer by Goldstern for Existence of unknowable algorithms ?

2013-04-05T15:20:01Z

Take any open question Q. ("Is the Riemann hypothesis true"? "Is P=NP?" etc)

There is an algorithm that will answer Q correctly.

It is trivial to prove (in classical logic) that one of the following programs works:

Program 1: Print "yes".
Program 2: Print "no".

Answer by Goldstern for Hereditarily Countable Names and Proper Forcing

2013-04-03T12:12:58Z

Counterexample: Let $\kappa $ be uncountable and let $\mathbb P= \kappa$ be an antichain (with a special weakest element $0_{\mathbb P}$, and let $\mathbb Q$ be forced to be the same forcing. Let $\sigma$ be the $\mathbb P$-name of the generic element of $\mathbb P$.

Now note that each "hereditarily countable" name uses (hereditarily) only countably many of the names $\check \alpha$, $\alpha<\kappa$. More precisely: For every subset $B\subseteq \{ \alpha: \alpha < \kappa\}$ define the family of HC-$B$-names naturally. (That is: all conditions and names appearing in the recursive construction must be in $B$, or HC-$B$-names, respectively.)

Then show that the set of HC names which are an HC-$B$-name for some countable $B$ is closed under the two operations (1) and (2).

Every HC-$B$-name for a condition is forced to be equal to the empty condition, or in $B$.

If the empty condition forces $\tau\le \sigma$ for some HC-$B$-name $\tau$, for some countable $B$. Now choose a condition $p_0\notin B$; then $p_0$ forces that $\sigma=p_0$, but cannot force $\tau=p_0$.

(I think the point of HC-names is this: If you have a countable set of names $\sigma_n$, and a condition $p$, then you can find a stronger condition $q$ and a set of HC names $\tau_n$, such that $q$ forces $\tau_n \le \sigma_n$ for all $n$. To prove this, let $N$ be a sufficiently large countable elementary model, and let $q$ be generic for $N$. Obtain $\tau_n$ from $\sigma_n$ by ignoring elements outside $N$.)

Answer by Goldstern for Bijection from $\mathbb{R}$ to $\mathbb{R^2}$

2013-03-31T12:15:18Z

Here is a bijection that uses the decimal (or binary, whatever) expansion of reals. (Even though I think the approach using continued fractions is more canonical.)

Let $\alpha:\mathbb Z\times \mathbb Z\to \mathbb N$ and $\beta:[0,1)\times [0,1)\to [0,1)$ be bijections.
Let $f:\mathbb R \to \mathbb Z \times [0,1) $ be the natural bijection $x\mapsto (\lfloor x\rfloor, x-\lfloor x\rfloor)$.
Together, $f$, $\alpha$, $\beta$ will define bijections $$ \mathbb R\times \mathbb R \to^f (\mathbb Z\times [0,1)) \times (\mathbb Z\times [0,1)) \simeq \mathbb Z^2 \times [0,1)^2 \to ^{\alpha,\beta} \mathbb Z \times [0,1) \to^{f^{-1}}\mathbb R$$

The main part is of course the definition of $\beta$. [EDIT: This is not my construction; I am not sure where I first read it. In his book on the real numbers, Oliver Deiser gives a very similar construction (blocking zeroes instead of nines) and calls this Julius König's trick. König's wikipedia page mentions it but omits the details.]

Represent each real number $x\in [0,1)$ as a sequence of DIGITS, where each DIGIT is either in $\{0,\ldots,8\}$ or is of the form $10*(10^k-1)+i$ with $k\ge 1$ and $i\in \{0,\ldots,8\}$ (i.e., in $\{90,\ldots, 98; 990,\ldots, 998; 9990,\ldots, 9998; \ldots\}$ .

For example, the number 0.0129449956$\dots$ would be represented by $(0,1,2,94,4,995,6,\ldots)$.

Given a pair $(x,y)$, $\beta$ interleaves these two representations.

Answer by Goldstern for Provability in Second-Order Arithmetic without the Successor Axiom

2013-01-27T13:12:11Z

This is not an answer, just an attempt at explicitly writing down my interpretation of the question. Too long for a comment.

Z2 and FPA are many-sorted FIRST ORDER theories, with one "lowercase" sort (for numbers) and infinitely many "uppercase" sorts (the n-th sort is for n-ary relations. There are predicates $Succ(x,y)$, $Add(x,y,z)$, $Mult(x,y,z)$ and $Element_n(x_1,\ldots, x_n,Y)$ (one for each $n$). However, we write $S(x)\sim y$ instead of $Succ(x,y)$; similarly $x+y\sim z$ instead of $Add(x,y,z)$, etc. Also we write $Y(x_1,\ldots, x_n)$ instead of $Element_n(x_1,\ldots, x_n,y)$. There is also a relation $\le$.

There are several groups of axioms. The first group says that successor, addition and multiplication are partial functions. The second group says that they satisfy the usual properties from Peano arithmetic whenever defined, e.g.

if $x+y\sim z$ and $S(z)\sim z'$ and $S(y)\sim y'$, then $x+y'\sim z'$.

I am not sure about the axioms for $\le$: Perhaps $0\le x$, and $S(y)\sim y' \to (x\le y' \leftrightarrow x\le y \vee x=y')$? Perhaps an axiom demanding that $\le$ is a total order? A discrete total order? (Feel free to edit.) And: "the domain of $S$ is downward closed".

In addition, Z2 (but not FPA) says that all functions are total.

The third group are the comprehension axioms: For every formula $\varphi(x_1,\ldots, x_n,\bar p)$ (where the parameters $\bar p$ may use come from all sorts), the universal quantification of $$\exists Y \ \forall x_1,\ldots, x_n \ [ Y(x_1,\ldots, x_n) \leftrightarrow \varphi(x_1,\ldots, x_n,\bar p)]$$

The fourth group is the induction axiom: $$ \forall Z: \quad [Z(0)\wedge \forall x \forall y\ ((Z(x)\wedge S(x)\sim y )\to Z(y))]\quad \to \quad \forall x \ Z(x) $$

The question is (correct me if I am wrong):

Assume that $\varphi$ is a formula (using any sorts) which is first order provable from Z2, and true in all standard models of FPA. Is $\varphi$ then first order provable from FPA?

(The standard models of FPA are: the natural numbers, and all truncated models. In all standard models, the uppercase sorts are interpreted as the respective power sets.)

Answer by Goldstern for Examples of "exotic" induction

2013-01-21T19:45:35Z

The Ackermann function is well-defined.

This is easily proved by lexicographic induction on $\mathbb N\times \mathbb N$. But perhaps not easily formulated -- I can only think of this awkward formulation: For every $x,y$ there is a unique function on the initial segment determined by $(x,y)$ satisfying the Ackermann recursion.

Finite support iterations of $\sigma$-centered forcing notions

2011-12-22T21:57:53Z

I am looking for a proof (or better, a reference) of the following fact:

The finite support iteration of $\sigma$-centered forcing notions is again $\sigma$-centered, assuming we iterate less than $(2^{\aleph_0})^+$ steps.

(EDIT: In the first version of the question I forgot to mention that it was Stefan Geschke who suggested that there should be a proof similar to "the product of continuum many separable spaces is still separable")

EDIT: In his 1994 paper "$\sigma$-centred forcing and reflection of (sub)metrizability" in TAMS (MR1179593 94g:54003), Frank Tall writes:

"It is well known (proved by the same method that proves the product of $\le 2^{\aleph_0}$ separable spaces is separable) that the finite support iteration of $\le 2^{\aleph_0}$ $\sigma$-centred orders is $\sigma$-centred."

Answer by Goldstern for Compactness of the Hilbert cube without the Axiom of Choice

2013-01-19T21:54:49Z

The compactness of the Hilbert cube follows (without AC) from the compactness of $2^\omega$, since $[0,1]$ as well as $[0,1]^\omega$ are continuous images of $2^\omega$.

(Conversely, $2^\omega$ is a closed subset of the Hilbert cube.)

The compactness of $2^\omega$ is just König's lemma for trees of binary sequences, which is easy to prove (hence certainly well-known) without AC. (I think this is called "weak König's lemma", an important principle in reverse mathematics.)

Answer by Goldstern for The canonical forcing of the GCH and direct limits.

2013-01-14T20:40:50Z

Let me take the case of $\aleph_\omega$ as a specific example. Similar considerations apply at other singulars.

If you want $\sigma$-closure above $\omega_1$, you need the full support at $\aleph_\omega$; bounded support (i.e., direct limit) will not be closed enough.

Also: You need not worry about collapsing $\aleph_\omega$ (or changing its cofinality). If the elements of an unbounded set of cardinals below $\aleph_\omega$ stay cardinals, then $\aleph_\omega$ will stay a cardinal as well (and if a cobounded set of cardinals below $\aleph_\omega$ is collapsed to $\lambda$, then $\aleph_\omega$ will certainly not become $\lambda^+$ in the extension, and hence collapse as well. If $\kappa$ is a singular cardinal of cofinality $\lambda < \kappa$, then the cofinality of $\kappa$ will only change if the cofinality of $\lambda$ changes. This will be true no matter what support you take.

Answer by Goldstern for Small Implications of the Axiom of Replacement

2013-01-13T16:05:50Z

The set $M$ of all formulas $\varphi$ that are of the form $V_{\omega+\omega} \vDash \psi$ is certainly recursive. Now the set $N:=\{ \varphi\in M: ZFC \vdash \varphi\}$ is c.e.

A standard trick gives an equivalent set $N'$ which is recursive (decidable): replace the $n$-th formula in $N$ (in any computable enumeration) by an equivalent formula that is much longer that all previous ones.

Is this set $N'$ what you are looking for? I realize that it does not have the nice form you probably wanted.

Answer by Goldstern for what axioms are between AC and Countable choice !

2013-01-04T20:47:31Z

Dependent choice, for example. Or choice for well-ordered families, see http://mathoverflow.net/questions/118060

Answer by Goldstern for Math for a cake

2012-12-29T18:46:36Z

Gödels incompleteness theorem in the language of modal logic (where $\Box\varphi$ means that $\varphi$ is provable - say in Peano Arithmetic - and $\bot=\lnot \top$ is any false statement): $$\Box \lnot \Box \bot \Rightarrow \Box \bot.$$

Answer by Goldstern for Math for a cake

2012-12-29T18:39:27Z

Gödel's completeness theorem: A (first order) sentence $\varphi$ is provable from the axioms $\Sigma$ iff it holds in every model of $\Sigma$: $$ \Sigma \vdash \varphi \Leftrightarrow \Sigma \vDash \varphi$$

Answer by Goldstern for Old books still used

2012-12-28T17:54:47Z

van der Waerden's Moderne Algebra was first published in 1930, I think. I use the book occasionally for my course, but am not sure which edition.

Answer by Goldstern for Cardinality of the set of maximal ideals in a Boolean ring/algebra

2012-12-22T10:21:39Z

This particular question is easy to answer. Many related questions about the relationships between various cardinals associated with Boolean algebra (or Boolean rings), such as: cardinality, number of ultrafilters, density, cellularity, distributivity etc, can be found in

Monk, J. Donald: Cardinal invariants on Boolean algebras. Progress in Mathematics, 142. Birkhäuser Verlag, Basel, 1996. 3-7643-5402-X. MR1393943

Answer by Goldstern for How would set theory research be affected by using ETCS instead of ZFC?

2012-12-20T13:10:20Z

There are several relatives (typically: subtheories and supertheories) of ZFC that are used in set-theoretic research. If somebody wants to do set theory based on ETCS or a related base theory, these theories would have to be translated to this new base (or, in the case of theories that are not so canonical, such as ZFC*, substitutes would have to be found).

Such a translation seems to be rather straightforward in the case of supertheories; a good translation would of course use the idioms of the target language.

Examples of supertheories:
- ZFC plus large cardinals
- ZFC plus definable determinacy (e.g., projective determinacy, or some consequences thereof - related to large cardinals).
- ZF plus AD, or ZF + $V=L(\mathbb R)$ (supertheory of ZF only)
- ZF(C) plus "V is a certain inner model". Weaker versions include:
  - ZFC plus cardinal arithmetic assumptions (GCH, SCH)
  - ZFC plus combinatorial principles ($\diamondsuit$, etc)
- ZFC plus forcing axioms (MA, PFA etc)
- others, including combinations (conjunctions) of the above
Examples of subtheories
- KP and related theories, which do not have full comprehension. (I think they are usually associated with proof theory rather than set theory)
- ZFC minus infinity (this is more closely associated with arithmetic than with set theory)
- ZF, or ZF plus weak versions of choice
- ZFC minus power set (plus instances of power set). Typical models are of the form $H(\chi)$.
- ZFC minus replacement (plus finitely many instance of replacement). Typical models are $V_\delta$.
- ZFC*, an often unspecified finite subset of ZFC, used to get around the "undefinability of truth", or to apply the reflection theorem. (Morally the same as the previous item.)
- ZFC minus Foundation, reflecting the fact that Foundation is hardly used outside set theory.
- Others, including combinations (intersections) of the above
Other relatives:
- ZF(C) with atoms, perhaps closer to mathematical practice than ZFC itself.
- NBG. The relation between NBG and ZFC is very well understood, as are the advantages and disadvantages: On the plus side, NBG can naturally talk about classes, and is finitely axiomatized (which might make it more amenable to automated theorem proving). On the other hand, the fact that not every subclass of $\mathbb N$ is a set can be inconvenient.
- MK and others.
- NF and NFU, sometimes claimed to be more natural than ZFC. While ZFC makes it awkward to talk about classes, NFU has problems with the function $x\mapsto \{x\}$ .

Answer by Goldstern for Construct a fixed-point set operator

2012-12-09T11:50:45Z

Obviously you mean "for every nonempty $T$". From a choice function one can construct a well-order of $S$, so "constructing" a choice function is the same as "constructing" a well-order.

Use the set $\omega_1$, the set of countable ordinals. It is well-ordered by $\in$.

Alternatively, consider the set $W$ of all well-ordered subsets of the rational numbers $\mathbb Q$. Define two elements of $W$ as equivalent if there is an order isomorphism between them. Divide $W$ by this equivalence relation and you get a well-ordered set.

Answer by Goldstern for Intersection of an uncountable number of sets.

2012-12-08T22:49:26Z

There is a quite general case when such an uncountable intersection is measurable: Let $I$ be the set of real numbers, and assume that not only every $E_i$ is measurable, but that these sets are Borel, and that moreover the set $$E:= \{ (x,i): x \in E_i \}$$ is a Borel set. Then the intersection of all $E_i$ is co-analytic (also called $\bf \Pi^1_1$) and hence measurable by a classical theorem. ("Borel" can be replaced by "co-analytic", but not by "measurable".)
It has already been pointed out that if $I$ is has sufficiently small cardinality then the intersection still has full measure, where "sufficiently small" depends on the underlying set theory.
But there is an interesting case where $I$ has size continuum. Assume that $I= \omega^\omega $ is the set of all functions from the natural numbers to the natural numbers. Assume (as above) that $E $ is a Borel or $\Pi^1_1$ set. Assume also that the map $i \mapsto E_i$ is decreasing in the following sense:

Whenever $i,j\in \omega^\omega$ and $i(n)\le j(n)$ for all $n$, then $E_i \supseteq E_j$.

Then the intersection of all $E_i$ is not only measurable but has full measure.

Answer by Goldstern for Cardinality of Equivalence Relation of Eventually Sublinear Functions

2012-12-07T11:44:37Z

(EDIT: I was busy writing this answer when Andreas Blass already submitted his answer. No need to read this answer, it is surprisingly similar to Andreas' answer.)

Every function $f: \mathbb R_0^+\to \mathbb R_0^+$ is completely determined by its restriction to $\mathbb Q \cup D(f)$, where $D(f)$ is the set of points where $f$ is not continuous. If $f$ satisfies $x\le y \Rightarrow f(x) \le f(y)$, then $D(f)$ is countable since the intervals $(\lim_{x\to a-} f(x), \lim_{x\to a+} f(x))$ are disjoint for $a\in D(f)$. So the number of nondecreasing functions is $\beth_1=2^{\aleph_0}$.

If the functions are not required to be nondecreasing, then there are $\beth_2 = 2^{\beth_1}$ equivalence classes, which can be seen as follows:

For every set $A \subseteq (0,1)$ define $f_A$ by $f_A(n+x) = \log(n)$ if $x\in A$, $n\in \mathbb N$. Let $f_A(n+x) = 0$ for $x\in [0,1]\setminus A$.

Answer by Goldstern for If d/dx is an operator, on what does it operate?

2012-12-05T11:54:22Z

I repeat (a variant of) my comment, even though I agree that it is shallow and has low entertainment value.

As long as we are only looking at functions in one variable, there is only one differential operator $D$, which may be called $\frac d{dx}$ or $\frac{d}{dt}$ depending on the context.

If you look at a composite function $f \circ g$, you may introduce the notation/abbreviation $x=g(t)$, $y=f(x)$, then

$\frac {d}{dx} f$ or $\frac d {dx} y$ is just $D(f)$,
and by $\frac{d}{dt} f$ or $\frac d{dt} y$ you mean $D(f\circ g)$.

So here both $\frac {d}{dx} $ and $\frac {d}{dt} $ have a meaning, and the meaning is different.

When we look at functions in, say, two variables (do they appear in first year calculus?), we implicitly introduce an (arbitrary) order of variables, say x is the first and t the second, and $\frac{\partial}{\partial x}$ is the partial derivative with respect to the first variable. This makes sense even if you treat functions as "variable-agnostic" sets of ordered pairs. (Which I do all the time, and do not find peculiar at all. Tastes differ.)

Of course, the intended meaning always depends on the context. If $f$ is a binary function, $\frac d {dt} f$ may be a variant notation for $\frac{\partial}{\partial t}f$, or it may be understood that we are really looking at a unary function $\hat f$ obtained by composing $f$ with some function $t \mapsto (x(t), y(t))$.

Answer by Goldstern for Why does the Solovay-Tennenbaum theorem work?

2012-12-02T23:57:46Z

There are $\kappa$ many stages in which you add Cohen reals. so you will have at least $\kappa$ reals at the end. This is really a very simple principle: "if you want to add something, add it."

THe other direction is slightly more involved: Do not add anything you don't want to add, and keep your fingers crossed that unwanted objects (such as: more than $\kappa$ reals) don't find their way into the construction anyway.

The "nice names" argument is very basic and practically always used whenever you want to get an upper bound on the number of new reals. Every new real $x$ is described by a family of countably many maximal antichains, plus a coloring functions mapping each antichain into $\{0,1\}$ --- assigning to condition $p$ in antichain $A_n$ the value that $p$ forces to $x(n)$. Count how many such families of colored antichains there are, this will be an upper bound on the continuum in the extension.

If you want to count subsets of $\lambda$, you will of course have $\lambda$ many antichains.

Answer by Goldstern for What notion captures the 'class' of all classes?

2012-12-01T22:28:37Z

To give a very specific (quite formalistic, and possibly very wrong - depending on your or even my beliefs) answer to some of your questions:

In ZFC there is no set that is the set of all sets, for this we introduce the notion of class.

I don't, because I use ZFC. Whenever I say "class", I mean "formula". (Today. I may change my mind tomorrow.) You may use NBG, of course.

But then what is the 'class' of all classes.

No such thing, in ZFC. (Well, there is the set of all formulas. But that is not what you mean.) No such thing in NBG either. Try KM.

Do we apply the same idea again? But then at what stage do we stop?

It depends on what you mean by "we". I stopped at ZFC. You may go as far as you want. You may even use a type-theoretic approach, in which there are infinitely many levels of this hierarchy. However, once you have countably many levels, you may ask how many levels there are. But now the pictures looks somewhat similar to a universe $V_{\delta+\omega}$, which ZFC handles very well.

I seem to recall that Fraenkel-Bar Hillel-Levy, "Foundations of Set Theory", has an enlightening and more detailed discussion of this topic.

Answer by Goldstern for How to calculate a power of a sum of ordinals (ordinal binomial theorem?)

2012-12-01T13:43:38Z

$\omega^{\alpha\cdot \omega^\gamma} = (\omega^\alpha )^{\omega^\gamma} \le
(\omega^\alpha + \cdots )^{\omega^\gamma} \le (\omega^{\alpha+1})^{\omega^\gamma} = \omega^{(\alpha+1)\cdot \omega^\gamma}$.

Wlog $\alpha,\gamma > 0$. Note that $2\cdot \omega^\gamma = \omega^\gamma$ for all $\gamma>0$, by induction and the associative law.

Now $(\alpha+1) \omega^\gamma \le \alpha\cdot 2 \cdot \omega^\gamma = \alpha\cdot \omega^\gamma$, and the result is $\omega^{\alpha \cdot \omega^\gamma}$.

Answer by Goldstern for Well-ordering with a topological property

2012-11-26T23:55:00Z

No. There cannot be a strictly increasing $\omega_1$-sequence of closed sets in a topological space with a countable base. Their complements would be unions of open sets from the basis, and it is impossible to drop elements of a countable set uncountably many times.

Answer by Goldstern for special extremally disconnected spaces with only finite isolated points

2012-11-22T23:31:45Z

Let $\kappa$ be a measurable cardinal with $\sigma$-complete ultrafilter $U$. Let $X$ be the set of all finite sequences from $\kappa$. For $s\in X$, $i\in \kappa$, we write $(s,i)$ for the sequence you get by appending $i$ to $s$, similarly $(s,i,j)$, etc. We call a subset $A \subseteq X$ closed if it has the following property:

Whenever $s$ in $X$, and almost all successors of $s$ are in $A$, then also $s$ is in $A$:

More precisely: If the set $\{i \in \kappa: (s,i)\in A\}$ is in $U$, then also $s\in A$.

[EDITED:] In other words: A set $O$ is open if for all $s\in O$ also almost all successors of $s$ are in $O$. (Using the countable completeness of $U$, a neighborhood base of $s$ is given by the sets $O_{s,F}:=\{s\}\cup \{(s,i,j,\ldots, k): i,j,\ldots ,k\in F\}$ , for $F\in U$. Using the fact that $U$ is non-principal one can show that these sets are clopen.)(DELETED, see Joseph's comment below.)

We check that $X$ is extremally disconnected: If $O$ is open, and $A$ is the closure of $O$, we claim that $A$ is open. So let $s\in A$. If $s\in O$, then $s$ has a neighborhood in $A$, and we are done. So assume that $s$ is not in $O$. Then almost all successors of $s$ must also be in $A$, otherwise $A\setminus \{s\}$ is closed. So $A$ is open.

The fact that $X$ is a p-space follows from the countable closure of $U$. It is clear that $X$ has no isolated points.

Answer by Goldstern for Examples where adding complexity made a problem simpler

2012-11-24T23:20:17Z

The complex object $\mathbb C$ is in some respects easier to understand that the apparently simpler object $\mathbb R$. (E.g., if we consider zeroes of polynomials, or the behavior of power series.)

Comment by Goldstern

2013-05-24T19:37:25Z

Certainly $\mathfrak b$, the unbounding number, can be embedded: Use functions on $\omega$, and interpolate them linearly. Also $\mathfrak b+1$ can be embedded, since for the embedding of $\mathfrak b$ we may use functions from $\omega$ into `$\{-1, -\frac12, -\frac13, \}$ etc. This can be continued.. At the moment I don't see if this gets us to everything below $\mathfrak b^+$, however.

Comment by Goldstern

2013-05-20T20:22:21Z

The smallest cardinal, really. (Assuming AC, of course. Otherwise there may not be such a cardinal.)

Comment by Goldstern

2013-05-20T20:19:59Z

Let $A$ and $B$ be equivalence classes of Cauchy sequences. Define $C$ as the set of all Cauchy sequences $(z_n)$ such that for all $(x_n)\in A$, all $(y_n) \in B$, the sequence $(x_n+y_n-z_n)$ converges to $0$. Isn't then $C$ the sum of $A$ and $B$?

Comment by Goldstern

2013-05-20T17:19:16Z

Assume without loss of generality that the leading coefficient is 1. Then a sufficient condition (which applies in your example) is certainly that $f(0,n)<0$. This only involves the constant coefficient. Another sufficient condition is that $f(2013)<0$. Another sufficient condition is that $f(r)<0$, where $r$ the geometric mean of the absolute values of the second and third coefficient. Etc, etc. None of these conditions is necessary, however. (This is why this is a comment, not an answer.)

Comment by Goldstern

2013-05-14T18:02:32Z

Use the binomial theorem on $(z-1)^y-1$. This question does not fit here.

Comment by Goldstern

2013-05-10T22:03:27Z

@ramiro: I think that a forcing notion is called "well-met" if any two conditions with a common lower bound have a greatest lower bound. This is slightly stronger than "DUB". jstor.org/stable/2274204

Comment by Goldstern

2013-05-09T21:47:16Z

I am sorry, my fingers must have slipped. What I want was: already ZF minus FOUNDATION shows that the complements of most proper classes are again proper classes. (And replacement is important in those proofs. For example, the class of even ordinals is a proper class, because using replacement you can get from it the class of all ordinals, which is not a set, by Burali-Forti. Or: the class of all singletons is a proper class, because by replacement you can get from it the class of all sets, which not a set, by Russell.)

Comment by Goldstern

2013-05-08T22:05:39Z

No, I think your problem is with replacement. Already ZF minus replacement shows that the complements of most proper classes are themselves proper classes. ("most" in a non-technical sense)

Comment by Goldstern

2013-05-08T05:21:13Z

We want $3^{n+k}=3^n\cdot 3^k$ to hold even if $n=0$.

Comment by Goldstern

2013-05-07T08:23:20Z

I agree with your comment to Asaf's answer that you did not formulate your question clearly enough. Or, better: the question you have in mind is not the one that you wrote. In my opinion, your question does not fit the title, because there is nothing "paradoxical" about having a proper class whose complement is also a proper class; compare this with the existence of an infinite set of natural numbers whose complement is also infinite.

Comment by Goldstern

2013-05-06T22:16:10Z

correct, thank you.

Comment by Goldstern

2013-05-03T22:47:51Z

Which theorem??? Please read the FAQ and then reformulate (or delete) your question.

Comment by Goldstern

2013-04-28T23:19:44Z

I think the usual name for this kind of definition is "recursive", not "implicit". An implicit definition would be of the form $f(\lambda_n)=0$, or $f(\lambda_n, \lambda_{n+1})=0$, etc. Each term in your sequence is defined quite explicitly as a function of the previous terms. Or did I miss something?

Comment by Goldstern

2013-04-27T17:13:55Z

It is a precise question, but I don't think it is appropriate for mathoverflow.

Comment by Goldstern

2013-04-27T10:41:30Z

Could it be that $(f_n)_{n=0}^\infty$ is a sequence of functions from $\ell_2$ into $\mathbb R$?

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