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3 votes
1 answer
415 views

Suslin lines hereditarily Lindelof

I need to prove that every suslin line is hereditarily Lindelof. Any idea will be helpfull.
Anusk's user avatar
  • 31
14 votes
3 answers
1k views

Order homomorphism functions on $\omega_1$

I posted the following question more than two years ago on MO (and then reposted on MSE), but the answer remains incomplete, so I thought I would rephrase it a bit (to make the statement clearer) and ...
Mirko's user avatar
  • 1,375
18 votes
2 answers
630 views

Is the notion of fixed point property for topological spaces an absolute notion?

Recall that a topological space $X$ has the fixed point property (FPP) if any continuous function $f: X\to X$ has a fixed point. Is the notion of FPP for topological spaces an absolute notion? More ...
Mohammad Golshani's user avatar
8 votes
1 answer
451 views

Strong measure zero sets and selection principles

A set of reals $X$ is strong measure zero if for any sequence of positive real numbers $ ( \epsilon_n ) _{n \in \omega }$ there is a sequence of open intervals $ ( a_n ) _{n \in \omega }$ which ...
Student's user avatar
  • 213
11 votes
1 answer
769 views

Is there a suitably generalized Baire property for topological spaces of arbitrary cardinalities?

Is there some suitable generalization to the notion of Baire property for topological spaces of arbitrary cardinalities which satisfies the following condition: The meager sets are sets which are ...
user38200's user avatar
  • 1,416
20 votes
2 answers
1k views

An order type $\tau$ equal to its power $\tau^n, n>2$

(This is a re-post of my old unanswered question from Math.SE) For purposes of this question, let's concern ourselves only with linear (but not necessarily well-founded) order types. Recall that: $...
Vladimir Reshetnikov's user avatar
2 votes
1 answer
173 views

Is there a maximal (or maximal Tychonoff) non normal space?

Is there a maximal (or maximal Tychonoff) non normal space? In "A Problem of Set-Teoretic Topology" the existence of a maximal Tychonoff space is asserted. Also there exists a perfectly normal maximal ...
Vahideh Bagheri's user avatar
17 votes
2 answers
905 views

Intersection of compact sets in the unit interval

Let $\mathscr K$ be an uncountable set such that every $K\in\mathscr K$ is a compact subset of $[0,1]$ with positive Lebesgue measure. Does it then follow that there exists an uncountable $\mathscr A\...
TaQ's user avatar
  • 3,584
4 votes
2 answers
403 views

Is there a Tychonoff space $X$ of cardinality not of the form $2^\alpha$ such that $|C(X)| = |X|$

Let $X$ be the real line with the usual topology. Then clearly $|C(X)| = c = |X|$ and on the other hand $|X| = 2^{\aleph_0}$. Now my question is as in the title: Is there a Tychonoff space $X$ of ...
user avatar
9 votes
1 answer
691 views

What is the shape of mathematical universe?

Shape? At the usual mathematical literature when we can discuss about the shape of a "space" that we have a kind of "topography" on it. For example a topology, metric, geometry, etc. Note that for ...
user avatar
3 votes
2 answers
432 views

When is a filter generated by a (countable) chain?

In any partial order $(P,\leq)$ it is easy to see that every chain generates (i.e., by taking the upwards closure) a filter, and any filter built as a result of the Rasiowa-Sikorski lemma in forcing ...
Iian Smythe's user avatar
  • 3,115
13 votes
1 answer
558 views

Idempotent ultrafilters and the Rudin-Keisler ordering

Short version: what can we say about the place of idempotent ultrafilters in the Rudin-Keisler ordering? Longer version: If $U$, $V$ are (nonprincipal) ultrafilters on $\omega$, then we write $U\ge_{...
Noah Schweber's user avatar
15 votes
1 answer
673 views

Question about product topology

Suppose $S\subset\mathbb{R}$ is dense without interior point, and for every open interval $I,J\subset\mathbb{R}$, $I\cap S$ is homeomorphic to $J\cap S$. Is $S\times S$ homeomorphic to $S$? By Luzin ...
duodaa's user avatar
  • 153
13 votes
2 answers
690 views

How many subsets of $\mathbb{R}$ are order isomorphic to $\mathbb{Q}$?

How many subsets of $\mathbb{R}$ are order isomorphic to $\mathbb{Q}$? How many subsets of the long line $\omega_1\times[0,1)$ are order isomorphic to $\mathbb{Q}$? I can see that results in both ...
Hanna K.'s user avatar
  • 233
11 votes
1 answer
799 views

Restrictions of null/meager ideal

Let I denote the null (resp. meager) ideal on reals. Is it consistent that for any pair of non null (resp. meager) sets A and B, there is a null (resp. meager) preserving bijection between A and B? In ...
Ashutosh's user avatar
  • 9,631
5 votes
1 answer
649 views

Showing a filter with a certain property on the power set of $\mathbb{Z}$ is a one point filter

Let $\mathcal{P}_0(X)$ the Power set of $X$ without the empty set and let $\dot{x}:=\{A\subseteq X: x \in A\}$ the one point filter generated by $x$. Furthermore let $$ \mathcal{A} := \{ f \in X^{\...
Dominic Michaelis's user avatar
10 votes
0 answers
314 views

How much do idempotent ultrafilters generate in terms of semigroups?

It is known that the set of ultrafilters on, say, the natural numbers $\mathbb{N}$, can naturally be endowed with the structure of a compact topological left semigroup (which fails to be anything ...
Jakub Konieczny's user avatar
9 votes
0 answers
361 views

Well-founded families of sets and topological convergence

Background/Motivation A space is scattered if every non-empty subset has an isolated point. A space is pseudoradial if every non-closed set contains a transfinite sequence (a well-ordered net) ...
Santi Spadaro's user avatar
6 votes
4 answers
473 views

Periodic point-free maps and free ultrafilters.

Let $X$ be a set and $u$ be a free ultrafilter on $X$. We can consider a topology on $X$ by declaring every element of $u \cup \{\emptyset \}$ to be open. El'kin's original motivation for looking at ...
Santi Spadaro's user avatar
10 votes
2 answers
750 views

Is there a compact space with no countably generated dense subspace?

This is a reformulation of this MO question which recieved little or no attention due to the fact that the OP gave no motivation whatsoever. I found the question quite interesting and decided to give ...
Ramiro de la Vega's user avatar
8 votes
3 answers
1k views

Axiom of Choice and continuous functions

Do you know if the following statement is an equivalent form of the axiom of choice or not? If $X$ is a compact metric space, then every continuous function $f: X \longrightarrow \mathbb{R}$ is ...
Ivan's user avatar
  • 249
4 votes
1 answer
252 views

A question on hereditary Lindelof number

Let $X$ be space. A space $X$ is called right-separated if it can be well-ordered in such a way that every initial segment is open in $X$. See the related link (left-separated). How could we show ...
Paul's user avatar
  • 654
4 votes
1 answer
255 views

Forcing over the poset of nonempty open subsets of a nice topological space

Is there anything sensible to be said concerning a notion of forcing given by the poset of nonempty open subsets of the sort of topological space that comes up in ($e.g.$ algebraic) topology? If so, ...
Adam Epstein's user avatar
  • 2,550
2 votes
0 answers
371 views

Descriptive set theory on $\mathbb{R}^\mathbb{N}$

The short version of my question is, What is a good source for learning about descriptive set theory on the space $\mathbb{R}^\mathbb{N}$, under the product topology coming from the discrete topology ...
Noah Schweber's user avatar
3 votes
1 answer
164 views

Algebras with countable chains only

Is there an example of an uncountable Boolean algebra $B$ in which every chain is countable and such that $\ell_\infty$ embeds into the Banach space $C(\mbox{Stone }B)$? The latter requirement is not ...
Bojan Kwitek's user avatar
1 vote
0 answers
321 views

Type I subspaces of the Stone Cech compactification of $\omega$

EDIT: I found a construction, see below. I decided not to delete the question in case someone is interested. A space $X$ is of Type I if $X=\cup_{\alpha<\omega_1} X_\alpha$, where each $X_\alpha$ ...
Mathieu Baillif's user avatar
6 votes
3 answers
281 views

Well-ordering with a topological property

Assuming the axiom of choice, is there a well-ordering of the reals such that every initial segment is closed for the usual topology? If the continuum hypothesis helps, we can also assume it. An ...
Denis's user avatar
  • 1,341
4 votes
1 answer
668 views

special extremally disconnected spaces with only finite isolated points

We Know that a cardinal $\kappa$ is measurable if there is a set $X$ with cardinal $\kappa$ and a {0,1}-measure $\mu: P(X) \rightarrow ${$0,1$} so that for all $x \in X$, $\mu(x)=0$ and $\mu(X)=1$. ...
Ali Reza's user avatar
  • 1,788
6 votes
1 answer
375 views

How much $\beta \mathbb{N}$ is homogenous?

Let $p,q\in \beta \mathbb{N}\setminus \mathbb{N}$. Must always the spaces $\beta \mathbb{N}\setminus \{p\}$ and $\beta \mathbb{N}\setminus \{q\}$ be homeomorphic? If no, can we for each point $p\in \...
Slavoj Žižek's user avatar
13 votes
1 answer
1k views

Is every $\sigma$-algebra of sets *abstractly* the Borel algebra of a topology on perhaps some other set?

Is every sigma-algebra the Borel algebra of a topology? inspires the present question which asks for less. Question: Given a $\sigma$-algebra $\mathcal A$ on a set $X$, does there exist a topology $\...
David Feldman's user avatar
7 votes
0 answers
466 views

Closure properties of familes of $G_\delta$ sets.

Given a family of sets $G\subset P(X)$, can one characterize by "closure properties" alone whether or not $G$ arises as the family of all $G_\delta$ for some topology on $X$? some Polish space ...
David Feldman's user avatar
6 votes
1 answer
634 views

Arbitrary small positive lower semi continuous functions

This question is a generalization of the question posed in this page to lower semi continuous functions. so let me describe the Question in the following way. Def: Let $(X,\tau)$ be a Tychonoff ...
Ali Reza's user avatar
  • 1,788
16 votes
1 answer
2k views

Characterization of Stone-Cech compactifications

Suppose I have an infinite discrete topological space $X$ of cardinality $\kappa$. Then I know some things about the Stone-Cech compactification, $\beta X$: it is Hausdorff and compact but not ...
Noah Schweber's user avatar
15 votes
1 answer
1k views

In ZF, when is a disjoint union of metrizable spaces metrizable?

It is easy to see that the disjoint union $\bigsqcup_i X_i$ of a collection of metric spaces is metrizable, simply by rescaling or chopping off the individual metrics to have diameter at most one, and ...
David Feldman's user avatar
0 votes
1 answer
278 views

On the compactness of a certain chain topology [closed]

Let $X$ be a non-empty set and $I$ a collection of some nested subsets of $X$ indexed by a linearly ordered set $(\Lambda,\le)$ such that $I$ always contains the void set $\emptyset$ and the whole set ...
K A Khan's user avatar
  • 243
16 votes
1 answer
1k views

Does Urysohn's Lemma imply Dependent Choice?

It's widely known$^{1}$ that in the proof of Urysohn's Lemma (UL) one uses the Principle of Dependent Choice (DC). Inspired by the equivalence between DC and Baire's Category Theorem$^{2}$, I'd like ...
Paulo Henrique's user avatar
6 votes
2 answers
257 views

Borel functions on $\omega_1$

Endow $\omega_1$ with order topology. It is easy to show that each continuous function $f\colon \omega_1\to \mathbb{R}$ is eventually constant. Is the same true for Borel functions?
Kulikov's user avatar
  • 61
92 votes
3 answers
14k views

Is every sigma-algebra the Borel algebra of a topology?

This question arises from the excellent question posed on math.SE by Salvo Tringali, namely, Correspondence between Borel algebras and topology. Since the question was not answered there after some ...
Joel David Hamkins's user avatar
6 votes
0 answers
561 views

Continuous images of Cantor cubes

The original title of this question was "Is there only one (up to homeomorphism) zero-dimensional homogeneous dyadic space of weight $\mathfrak{c}$?". I changed it with the hope of getting a bit more ...
Ramiro de la Vega's user avatar
2 votes
1 answer
220 views

Extending BAs to weakly countably distributive algebras.

Suppose $\mathscr{A}$ is a complete Boolean algebra (assume it is c.c.c. if you wish). Is there any, say, canonical embedding $\mathscr{A}\subseteq \mathscr{B}$ into a complete Boolean algebra which ...
TomK's user avatar
  • 55
2 votes
1 answer
274 views

Does X have any diagonal properties?

Assume that $2^{\omega_1}=2^\omega=\mathfrak{c}$. Let $D$={ 0,1 }, and let $Y=D^\mathfrak{c}$. For $y\in Y\;$ let $\operatorname{supp}(y)$={$\xi<\mathfrak{c}:y(\xi)=1$}, the support of $y$, and let ...
Paul's user avatar
  • 654
4 votes
1 answer
399 views

If a topological space X has $\aleph_1$-calibre, then it must be star countable?

If a topological space X has $\aleph_1$-calibre[definition], then it must be star countable? What if the cardinality of the topological space X is additionally < = $2^{\aleph_0}$?
Paul's user avatar
  • 654
3 votes
2 answers
510 views

Is there a countable pseudocharacter Hausdorff space,such that...?

Let X be a Hausdorff space and Difine the Property A as following: if $\mathscr{U}$ is a collection of open sets of X that witnesses Hausdorff property of X (= $\forall x,y \in X$, there exist two ...
Paul's user avatar
  • 654
7 votes
5 answers
1k views

the example of ccc but not separable

I am interested in the relation between the property of countable chain condition (ccc) and the property of separable. Could someone recommend some papers or books about this to me? thanks in advance.
Paul's user avatar
  • 654
0 votes
1 answer
501 views

$\aleph_1$-calibre

The square of X which is $\aleph_1$-calibre is still $\aleph_1$-calibre?
Paul's user avatar
  • 654
8 votes
1 answer
802 views

A topological space for which having the ccc is independent of ZFC?

It is well known that a generalized Cantor space $2^A$ is separable if and only if $|A| \leq 2^{\aleph_0}$. This means that one cannot decide in $ZFC$ whether the space $2^{\omega_2}$ is separable or ...
Ramiro de la Vega's user avatar
8 votes
1 answer
365 views

Counting copies of a BA within a BA: arbitrarily many vs infinitely many

Informally, I am wondering if a Boolean algebra $\mathcal{B}$ contains infinitely many disjoint copies of a Boolean algebra $\mathcal{A}$ whenever it contains arbitrarily many disjoint copies of $\...
Asher M. Kach's user avatar
7 votes
2 answers
597 views

A characterisation of well-ordering ?

It is easy to prove that if $E$ is well-ordered, and if $f$ is a strictly increasing map from $E$ to $E$, then, for all $x$ in $E$, $f(x) \ge x$ (just consider the sequence $x$, $f(x)$, $f(f(x))\dots$)...
Feldmann Denis's user avatar
23 votes
3 answers
2k views

An ultrafilter is a set of subsets containing exactly one element of each finite partition: reference request

There are probably dozens of ways of defining "ultrafilter". The definition I've seen most often involves first defining "filter", then declaring an ultrafilter to be a maximal filter. But there's ...
Tom Leinster's user avatar
  • 27.7k
1 vote
0 answers
150 views

Follow up question on the measure of the difference between a partial selector and a selector...

This is a different question from my previous question Difference between a partial selector and a selector, however I am going to repeat the preamble... In Kharazishvili's "Nonmeasurable Sets and ...
George Lazou's user avatar

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