gn.general topology – Does every open set contain a dense \$F_{sigma}\$ subset?

Let $$U$$ be a regular open set in a Tychonoff space $$X$$ (regular means that it is an interior of a closed set).

( In my specific situation $$U$$ is of the form $$int f^{-1}(0)$$, where $$f$$ is a continuous real-valued function on $$X$$, and $$X$$ is a Baire space (a sequence of dense open sets has a dense intersection), but I am not sure if it helps. )

Is there a sequence $${A_n}_{ninmathbb{N}}$$ of closed (in $$X$$) subsets of $$U$$ such that $$bigcup_{ninmathbb{N}} A_n$$ is dense in $$U$$?

Of course, this is the case if $$X$$ is perfectly normal (which is equivalent to every open set being $$F_{sigma}$$), or separable, but I hope a less restrictive assumption will suffice, e.g. normality.

gn.general topology – A question about locally compact spaces

Recently I read a book about linear algebraic group written by Ian Macdonald. There is a conclusion which I can’t prove.

It says that if $$X$$ is locally compact Hausdorff space, then $$X$$ is compact if and only if, for all locally compact spaces $$Y$$, the projection $$Xtimes Y to Y$$ is a closed map. Is it a fact for all topology spaces?

gn.general topology – Can Tychonoffs theorem for a countable number of spaces be proven with ZF plus the axiom of (countable) dependent choice?

It can be proven without any form of infinite choice that the product of two compact spaces (and thus any finite product) is compact, while on the other hand, it is well known that the general form of Tychonoffs theorem implies the axiom of choice.

So a general formulation of this question would be: Is there a proof of Tychonoffs theorem where the use of AC/Zorn’s lemma can be reduced to using the axiom of dependent choice up to the ordinal that indexes the product? With the case of countable products being of particular importance.

gn.general topology – How to construct this dendrite?

In the early 1970s Pelczynski noticed that the only surjective isometries on $$C(K)$$ for the following compact Hausdorff space $$K$$ are $$pm Id$$. I believe this was the first such example.

Quoting from Davis in “Separable Banach spaces with only trivial isometries”: Let $$K$$ be a dendrite containing for each $$n>2$$ exactly one cut-point of degree $$n$$ so that the cut-points are dense in $$K$$. Then the only surjective homomorphism of $$K$$ is the identity.” Banach-Stone then yields the result.

My question is: How does one construct $$K$$?

One of my issues is that I don’t know some of the basic definitions. What is the degree of a cut point? Googling has not been very helpful.

gn.general topology – Example of a compact \$alpha\$-normal space which is not normal

Could you help me to find an example of a compact space which is $$alpha$$-normal, but not normal? The definition of $$alpha$$-normal is any disjoint closed subsets $$A, B$$ can be separated by disjoint open subsets $$U, V$$ such that $$Ucap A$$ is dense in $$A$$ and $$Vcap B$$ is dense in $$B$$.

Thank you.

gn.general topology – Existence of a quasi-open (a.k.a semi-open) map into a Cantor cube

Recall that a topological space is extremally disconnected if the closure of any open set is open.

A continuous map is quasi-open if it maps open sets onto sets with nonempty interior. For some reason this class of maps shows up in the literature under different names, the most common of which is semi-open.

If $$K$$ is an extremally disconnected compact Hausdorff space without isolated points does the exist an infinite cardinal $$alpha$$ and a quasi-open map $$varphi:Kto{0,1}^alpha$$?

gn.general topology – Sequences with 0’s in \$mathbb R ^omega\$

Let $$mathbb R ^omega$$ be the set of all sequences of real numbers in the product topology.

Let $$X$$ be the set of all sequences in $$mathbb R ^omega$$ which have at least one 0.

Let $$Y$$ be the set of all sequences in $$mathbb R ^omega$$ which have at least two 0’s.

Are $$X$$ and $$Y$$ homeomorphic, and if so, is there a simple proof of this?

gn.general topology – If \$X^{star} times Y^{star}\$ is \$KC\$ space, then \$Xtimes Y\$ is a \$k\$-space

The following theorem is found in the article “ON KC AND k-SPACES, A. García Maynez. 15 No. 1 (1975) 33-50”

Theorem 3.5: Let $$X, Y$$ be topological spaces. If $$X^{star} times Y^{star}$$ is KC then $$Xtimes Y$$ is a $$k$$-space

Definitions:

1. $$X$$ is a $$KC$$ space if every $$Ksubset X$$ compact is closed.
2. Let $$A⊂X$$. Then $$A$$ is a $$k$$-closed if for all $$K⊂X$$ compact it happens that $$A∩K$$ is closed in $$K$$
3. $$X$$ is a $$k$$-space if every $$k$$-closed set of $$X$$ is a closed set in $$X$$.
4. Let $$Csubset X$$. Then $$C$$ is compactly closed if for all $$Ksubset X$$ compact and closed, $$Ccap K$$ is compact.

The article refers that the theorem 3.5 is a consequence of theorem 3.4

Theorem 3.4: Let $$X,Y$$ be non-compact spaces and let $$Y^{star}=Ycup { infty }$$ be the one-point compactification of $$Y$$. Assume $$Y$$ is a $$KC$$ space. Then a set $$Csubset Xtimes Y$$ is compactly closed in $$Xtimes Y$$ if and only if $$Ccup (Xtimes { infty })$$ is compactly closed in $$Xtimes Y^{star}$$.

My attempt:
Let $$Asubset Xtimes Y$$ $$k$$-closed. I want to show that $$A$$ is closed in $$Xtimes Y$$.

By hypothesis and $$KC$$ is a hereditary property, we have to $$Xtimes Y$$ is $$KC$$ space. Also $$KC$$ is a factorizable property, then $$Y$$ is a $$KC$$ space. Now, clearly every $$k$$-closed set is compactly closed, therefore $$A$$ is compactly closed in $$Xtimes Y$$. Then by the theorem 3.4 we have $$Acup (Xtimes {infty })$$ is compactly closed in $$Xtimes Y^{star}$$.

Let $$Ksubset Xtimes Y$$ compact and closed. Consider the projection function $$rho_2 : Xtimes Y rightarrow Y$$, which is continuous and by the compactness of $$K$$, then $$rho_2 (K)$$ is compact and closed in Y, beacuse $$Y$$ is $$KC$$. Note that $$Y-rho_2 (K)$$ is open in $$Y^{star}$$ and $$Y^{star}-rho_2 (K)=Y^{star}cap (Y-rho_2 (K))$$, which is open in $$Y^{star}$$, therefore $$rho_2 (K)$$ is closed in $$Y^{star}$$ and $$Xtimes rho_2 (K)$$ is closed in $$Xtimes Y^{star}$$. As $$Ksubset Xtimes rho_2 (K)$$ we have $$K$$ is closed in $$Xtimes Y^{star}$$

Now remember $$Acup (Xtimes {infty })$$ is compactly closed in $$Xtimes Y^{star}$$ it happens that $$Acap K=Kcap (Acup (Xtimes {infty }))$$ is compact in $$Xtimes Y$$. Therefore $$Acap K$$ is closed in $$Xtimes Y$$ because $$Xtimes Y$$ is KC.

I still don’t have a clear idea of how to conclude that $$A$$ is closed in $$Xtimes Y$$. I hope you can help me or make any observations of my proof

Thank you.

gn.general topology – When the union of open subsets of subsets is open

Let $$X$$ be a topological space and $${A_i}_{i in I}$$ be a family of closed subsets of $$X$$ such that for $$inot= j$$ we have $$A_icap A_j={}$$. Now for each $$iin I$$ let $$A’_i$$ be an open subset of $$A_i$$ when considering $$A_i$$ as a subspace of $$X$$, that is, $$A’_i=O_icap A_i$$ for some open subset $$O_i$$ of $$X$$. Under what conditions (on $$X$$ or $$A_i’s$$ or…) we can deduce $$cup_{iin I} A’_i$$ is an open subet of $$cup_{iin I} A_i$$ ?

gn.general topology – Hypotesis on \$W\$ to achieve \$d(x,F(W))

Writing a paper, I’m trying to formulate the following technical result:

Let $$X$$ be a manifold, $$Wsubsetsubset X$$.
Let $$f_k:U_kto X$$ be continuous, where $$U_ksubset X$$ is an open neighborhood of $$K_k$$ and $$K_1:=overline W,;;K_{k+1}:=f_k(K_k)$$.

Assume $$V_ksubset X$$ is open and $$epsilon_k>0$$ are such that
begin{align*} (i)&;V_ksubset U_k\ (ii)&;operatorname{diam} V_k

Assume $$F_k:=f_kcirccdotscirc f_1:Wto X$$ converges uniformly on compact subsets of $$W$$ to some $$F:Wto X$$.
Then,

if $$xin K_k$$, $$d(x,F(W))le2cdotsum_{nge k}epsilon_n$$
holds true.

(the notion of distance follows from the topology of $$X$$ or from some Riemannian metric, which induces the same topology).

If $$xin K_k$$, one can easily see (by splitting the cases $$xin K_kcap V_k$$ and $$xin K_ksetminus V_k$$ and exploiting $$(i)-(iv)$$) that
$$d(x,K_{k+m+1})le2cdotsum_{n=k}^mepsilon_n$$
hence it seems a reasonable result but I’m having some hard times proving it, in fact we can’t just pass to the limit in this last one, since
$$K_{k+m+1}=F_{k+m}(overline W)$$
and $${F_k}$$ converges on compacts of $$W$$, not on $$overline W$$.

I tried to consider $${C_j}_j$$, a (almost) normal exhaustion of $$W$$, that is
$$C_jsubset W;;mbox{compact},;;C_jsubsetneq C_{j+1},;;bigcup_jC_j=W$$
(the usual notion of normal exhaustion involves $$C_jsubsetoperatorname{int}C_{j+1}$$, but $$W$$ itself could have empty interior, hence this notion would be void; and what I wrote seems to be reasonable); assuming that such a sequence exists (it doesn’t look like a problem!), for every $$epsilon>0;exists j_{epsilon}$$ such that, setting $$K_k^{(j)}:=F_{k-1}(C_j)$$, one has
$$d(x,K_k^{(j)})
Let $$yin C_j$$ realizing the minimum:
$$d(x,K_k^{(j)})=d(x,F_{k-1}(y)).$$
then
begin{align*} d(x,F(W)) &le d(x,F(C_j))\ &le d(x,F(y))\ &le d(x,F_{k-1}(y))+sum_{nge k-1}d(F_{n+1}(y),F_n(y)) end{align*}
the last sum is finite because of normal convergence, but I don’t see how make it depend on the $$epsilon_k$$, for example, to get some control on it.

My goal is to prove that what happens at step $$k$$ is not modified too much when we keep going. For example if we fix $$alpha>0$$ small and define $$epsilon_k:=alpha/2^{k+1}$$ we would get that $$x$$ doesn’t move more than $$alpha$$ from the image $$F(W)$$.

Any hint? Do you think it’s a false statement (so I should modify the hypotesis a bit)?. Thank you very much