Weight of locally compact hausdorff space does not exceed its cardinality.

Question

This seems to be assumed in a proof I'm reading. Is it true? How can it be proved?

Answer 1

A network for a space $X$ is a collection $\mathcal{N}$ of subsets of $X$ such that every open subset of $X$ can be written as a union of members from $\mathcal{N}$ or equivalently (just as with bases): For all open sets $O$ and all $x \in O$ there is some $N \in \mathcal{N}$ such that $x \in N \subseteq O$.

So a network is like a base, but its members are not necessarily open. Like with weight, defined as the minimal infinite cardinality of a base, we define $nw(X) = \min\{|\mathcal{N}|: \mathcal{N} \text{ a network for } X\} + \aleph_0$, which is a well-defined cardinal number.

Note that for any $X$ the set $\{\{x\}: x \in X\}$ is a trivial network for $X$, so $nw(X) \le |X|$ and as a base is certainly a network too, $nw(X) \le w(X)$, both for any space. Also if $Y$ is a subspace of $X$, $nw(Y) \le nw(X)$, just as we have that $w(Y) \le w(X)$. Also it's quite easy to show that for metrisable spaces $X$ we have $w(X) = nw(X)$, but this also holds for another large class of spaces:

Theorem: Let $X$ be compact and Hausdorff then $nw(X) = w(X)$.

Proof: suppose $(X, \mathcal{T})$ has a network $\mathcal{N}$ of minimal size $\kappa$. Define the set $I$ as $$\{(N_1,N_2) \in \mathcal{N} \times \mathcal{N}: \exists O_1, O_2 \in \mathcal{T}: N_1 \subseteq O_1, N_2 \subseteq O_2, O_1 \cap O_2 = \emptyset\}$$

Then $I$ has size $\le \kappa^2 = \kappa$ and for each $i =(N_1, N_2) \in I$ pick $O_1(i),O_2(i) \in \mathcal{T}$ that fulfill the promise from the defining condition. (AC required).

Then let $\mathcal{T'}$ be the topology on $X$ generated by the collection of those $\{O_1(i), O_2(i): i\in I\}$, and note $(X, \mathcal{T}')$ has weight $\le \kappa$, as the generating set has at most that size.

Then $(X,\mathcal{T'})$ is Hausdorff: let $x \neq y$ in $X$. Then as $(X, \mathcal{T})$ is Hausdorff we have $O_1, O_2 \in \mathcal{T}$ with $x \in O_1$, $y \in O_2$ and $O_1 \cap O_2 = \emptyset$. As $\mathcal{N}$ is a network we have $N_1, N_2 \in \mathcal{N}$ such that $x \in N_1 \subseteq O_1$ and $y \in N_2 \subseteq O_2$. Then the $O_1$ and $O_2$ witness that $i_0 = (N_1,N_2) \in I$ and so we have $O_1(i_0), O_2(i_0) \in \mathcal{T}'$ (not equal to our current $O_1$ and $O_2$ necessarily, of course) such that also $x \in N_1 \subseteq O_1(i_0)$, $y \in N_2 \subseteq O_2(i_0)$, and $O_1(i_0) \cap O_2(i_0) = \emptyset$. So we have shown Hausdorffness (and hopefully you see why we defined $I$ this way: exactly to get a coarser Hausdorff topology!).

The last part is classical: $1_X: (X, \mathcal{T}) \to (X, \mathcal{T}')$ (the identity map) is continuous as $\mathcal{T}' \subseteq \mathcal{T}$ and as the right hand space is Hasudorff, as we saw, and the left hand space is compact, we have that $1_X$ is also closed and a closed continuous bijection is a homeomorphism. So in fact $\mathcal{T} = \mathcal{T}'$ and $w(X,\mathcal{T}) \le \kappa$. Q.E.D.

Now the OP's question is for locally compact Hausdorff spaces. These have a one-point (Aleksandrov) Hausdorff compactification $\alpha X$ (with $X \subseteq \alpha X$ and $|\alpha X \setminus X| \le 1$) and it's easy to see that $nw(X) = nw(\alpha X)$ (just add a singleton to a network for $X$ e.g.). So for locally compact Hausdorff spaces we have $nw(X) = nw(\alpha X) = w(\alpha X) \ge w(X) \ge nw(X)$ by the theorem for compacta and the general facts.

Corollary: let $X$ be locally compact Hausdorff, then $w(X) \le |X|$.

This follows from $w(X) = nw(X)$ and $nw(X) \le |X|$, of course.