5

Let $(X,\tau)$ be a topological space $T_3$. Show that the following statements are equivalent:

  1. Every open and finite coverage of X has a finite refinement consisting of connected sets.
  1. Space X is locally connected and countably compact.

A topological space is called countably compact if every open and enumerable coverage admits a finite subcoverage.

Any ideas:

$1 \to 2$ To prove that $X$ is locally connected I thought of the following, by hypothesis $A=\{X\}$ has a finite refinement consisting of connected sets, say $B=\{B_i:B_i \text{ is connected and } i \in \{1,...,n\}\}$. Let $x \in X$ and $U$ open of $X$, such that $x \in U \subset X$, since $X=\cup B_i$, $x \in B_k$ where $k \in \{1,...,n\}$, if $B_k \subset U$ and $B_k$ is open, $X$ is locally connected in $x$, If the above does not happen, I don't know how to proceed.

proving that it is countably compact does not occur to me how.

$2 \to 1$ let $U=\{U_i:i \in \{1,2,...,n\}\}$ be a finite coverage of $X$. Since $X$ is locally connected, for each $x \in U_i$ there is a connected open $V_{ix}$ such that $V_{ix} \subset U_i$. Thus, $U_i=\cup _{x \in U_i} V_{ix}$. My idea here was to use the hypothesis that $X$ is countably compact and apply it to each $U_i$, but I'm not sure of this fact, because I don't know if there is a countable amount of $V_{ix}$ that covers $U_i$.

I don't know how to use the hypothesis that $X$ is $T_3$.

any help would be very useful. Thank you

Alex Ravsky
  • 106,166
user1999
  • 504
  • Is the refinement supposed to consist of open sets? Do you also assume that the refinement is a cover? Otherwise $2$ implies $1$ just by local connectivity. – Keen-ameteur Oct 17 '20 at 08:20
  • 1
    @AlexRavsky Doesn't local connectivity imply that every $x\in U$ where $U$ is open, have an open connected subset $x\in V\subseteq U$? – Keen-ameteur Oct 17 '20 at 09:04

1 Answers1

1

You question is on the equivalence of conditions (a) and (c) of Theorem 3.1 from [HI].

References

[HI] Henriksen, M., Isbell, J. R. Local connectedness in the Stone-Čech compactification, Illinois J. Math. 1: 4 (1957), 574–582.

Alex Ravsky
  • 106,166
  • in this paper when they say that "Suppose next that X is not countably compact, and let $D ={d_i }$ denote a countably ,infinite closed discrete subset of X. " why does that make that relation of countably compact and a countably ,infinite closed discrete subset of X? is it some equivalence? Could you please give me an explanation? I would appreciate it – user1999 Oct 19 '20 at 10:50
  • 1
    @user1999 It is well-known easy to check that a space is $X$ countably compact iff each infinite subset $D$ of $X$ has an accumulation point $x\in X$ (the latter means that each neighborhood of $x$ contains infinitely many points of the set $D$). So if the space $X$ is not countably compact then it contains a countably infinite subset $D$ without accumulation points. If the space $X$ is $T_1$ then the set $D$ is closed and discrete. – Alex Ravsky Oct 19 '20 at 13:14