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.
 A: 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.
A: 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.
