# About Cardinality in Hausdorff Spaces

I have two problems:

1.- Let $X$ be a compact Hausdorff space, then $X$ has a basis with cardinality less than or equal to $|X|$.

2.- Let $X$ be a Hausdorff space and $D$ a dense subset in $X$, then $|X|\leq|P(P(D))|$, where $P(D)$ is the power set of $D$.

If somebody know where I can find the proofs, tell me please.

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All-caps in questions might be considered a bit rude. Also, is this homework? EDIT: Also also, I'm not sure I see two problems. –  Jack Henahan Jun 15 '11 at 1:45
I only see one question here. What is the other problem? –  Zev Chonoles Jun 15 '11 at 1:45
Thanks, @Zev, for editing in the questions. –  Jack Henahan Jun 15 '11 at 1:50
Should the first problem say "...less than or equal to $|X|$" or perhaps "... less than or equal to that of $X$"? –  Arturo Magidin Jun 15 '11 at 4:19

Problem 1. follows from the fact that the network weight of a compact Hausdorff space equals the weight of it, but this might be unfamiliar. A dedicated proof of this fact:

Let $\mathcal{T}$ be the original topology on $X$, the compact Hausdorff one. We can assume that $X$ is infinite, otherwise $X$ is discrete and finite, and its set of singletons is the required base.

For every pair $\{x,y\}$ with $x \neq y$ from $X$, we pick $U(x,y)$ and $V(x,y)$ from $\mathcal{T}$ such that $x \in U(x,y), y \in V(x,y), U(x,y) \cap V(x,y) = \emptyset$. Let $\mathcal{T}'$ be the topology generated by all sets $U(x,y)$ and $V(x,y)$, so that $\mathcal{T}'$ has a base of weight at most $X$ (as the cardinality of $X$ and the set of pairs from $X$ are equal, $X$ being infinite).

The identity map $f(x) = x$ from $(X,\mathcal{T})$ to $(X,\mathcal{T}')$ is a bijection that is continuous (as $\mathcal{T}' \subset \mathcal{T}$) and goes from a compact space to a Hausdorff one (we basically constructed the latter space to be Hausdorff), and so is a closed map as well, and hence a homeomorphism, from which we conclude that $\mathcal{T} = \mathcal{T}'$ and we are done.

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Let $A_q$ be the element in $2^{P(D)}$ which associates with each powerset $D$ a $1$ if the closure contains $q$, and $0$ otherwise. Then I claim that $A_q$ is unique for each point $q$ in the Hausdorff space, which proves the claim since $$|2^{P(D)}|=|P(P(D))|$$

If $A_p=A_q$, then consider disjoint neighborhoods $V_p$ and $V_q$ of $p$ and $q$ respectively. The closure of the elements of $D$ inside $V_p$ (which contains $p$ since every open set containing p must contain one of these points), cannot contain $q$ since it is contained in the closed set $\bar{V_q}$, so we have a contradiction!

Cheers,

Rofler

Edit: Sorry, I only saw your second question when you originally posted.

Edit2: Fixed legitimate concerns in my proof.

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Rofler: The interval $(0,1)$ and the open set $(0,1/\pi)\cup(1/\pi,1)$ contain the same rationals, so I am not sure I understand your first sentence. –  Andres Caicedo Jun 15 '11 at 2:10
I have edited my answer to reflect your legitimate concern that my original proof was total nonsense :D –  Rofler Jun 15 '11 at 6:37
More conventional rephrasing: every element of the closure of $D$ is the limit of a filter on $D$. How many filters are there on $D$? At most $2^{2^{|D|}}$. –  GEdgar Jun 15 '11 at 13:04
Sorry, I never learned about filters. I just blindly applied the definition of Hausdorff space :P –  Rofler Jun 15 '11 at 13:28