Size of topological space depending on the size of local basis. (With elementary submodels)

Question

Recall that the character of a topological space $\chi(X)$ is the minimum cardinal $\kappa$ such that every point in $X$ has a local basis of size $\kappa$.

I need to prove that if $X$ is compact $T_2$ topological space then $|X| \leq 2^{\chi(X)}$.

There have been similar results, perhaps the most known of them by Archangelskii.

Any pointers? Hints? Ideas? Most probably we will need to use elementary submodels of $V$ at some point.

Answer 1

Well, you could apply Arkhangel'skiĭ's theorem:

Theorem. If $X$ is Hausdorff, then $|X| \leq 2^{L(X) \cdot \chi(X)}$

where $L(X)$ denotes the Lindelöf degree of $X$, the smallest infinite cardinal $\kappa$ such that every open cover of $X$ has a subcover of size $\leq \kappa$.

I assume this isn't what you want. Below I attempt to give an sketch of a proof using elementary submodels.

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Let $\kappa = \chi (X)$, and fix a sufficiently large regular cardinal $\theta > 2^\kappa$. Let $M$ be an elementary submodel of $H(\theta)$ of size $\leq 2^\kappa$ containing (as elements) $X$, its topology, and $\alpha$ for each $\alpha \leq \kappa$. Furthermore, assume that $M$ has the property that if $A \subseteq M$ has cardinality $\leq \kappa$, then $A \in M$.

From here the proof goes via two claims:

Claim 1. $X \cap M$ is a closed subspace of $X$.

proof sketch. Let $x \in \overline{ X \cap M }$, and fix a local base $\mathcal{U} = \{ U_\alpha : \alpha < \kappa \}$ at $x$ of cardinality $\leq \kappa$. Use this to construct a net consisting of points of $X \cap M$ which converges to $x$, and argue that this net (or at least an equivalent one) belongs to $M$. By elementarity, $M$ thinks that there is a limit of this net, and by Hausdorffness the limit is unique. $\Box$

Claim 2. $X \subseteq M$.

proof sketch. Suppose that $x \in X \setminus M$. For each $y \in X \cap M$ we can fix a local base $\mathcal{U}_y \in M$ at $y$ of cardinality $\leq \kappa$, and note that $\mathcal{U}_y \subseteq M$. Now take some $U_y \in \mathcal{U}_y$ which does not contain $x$. Cover $X \cap M$ by sets in these families which do not contain $x$. By compactness some finite subcollection also covers $X \cap M$. Now note that $M$ thinks that this finite subcollection actually covers all of $X$. $\Box$