# Show that $sw(X) \leq nw(X)$

I'm reading through Hodel's paper on Cardinal Functions in "Handbook of Set-Theoretic Topology", and he states the following result on cardinal functions (I've added the definitions):

For any topological space $X$,

$sw(X)$ is the smallest (infinite) cardinal $\kappa$ such that there is a separating open cover of $X$ of cardinality $\kappa$, where a cover $\mathcal{V}$ is separating iff $\bigcap \{V \in \mathcal{V} \, : \, x \in V\} = \{x\}$ for all $x \in X$.

$nw(X)$ is the smallest (infinite) cardinal $\kappa$ such that there is a net of cardinality at most $\kappa$, where a collection $\mathcal{N}$ of subsets of $X$ is a net iff all open sets are unions of members of $\mathcal{N}$.

Then the result is that $sw(X) \leq nw(X)$ for $X$ a $T_2$ space.

I can't see why this holds. The proof is probably going to be some method of constructing a separating open cover from a given net. I have the fact that any net is a separating cover from the $T_2$ property, but this doesn't necessarily give me a separating open cover. Taking the interior of each member of the net doesn't work.

Let $\mathscr{N}$ be a net for $X$. Let $\mathscr{P}$ be the family of pairs $\{N_1,N_2\}\subseteq\mathscr{N}$ such that $N_1\cap N_2=\varnothing$, and there are disjoint open $V_1,V_2$ such that $N_1\subseteq V_1$ and $N_2\subseteq V_2$. For each $P=\{N_1,N_2\}\in\mathscr{P}$ fix such a pair $Q_P=\{V_1,V_2\}$ of open sets, and let $\mathscr{V}=\bigcup_{P\in\mathscr{P}}Q_P$; then $\mathscr{V}$ is a family of open sets, and $|\mathscr{V}|\le|\mathscr{N}|$.

Now let $x\in X$. $X$ is Hausdorff, so for each $y\in X\setminus\{x\}$ there are disjoint open sets $W_1$ and $W_2$ with $x\in W_1$ and $y\in W_2$. $\mathscr{N}$ is a net for $X$, so there are $N_1,N_2\in\mathscr{N}$ such that $x\in N_1\subseteq W_1$ and $y\in N_2\subseteq W_2$. Let $P=\{N_1,N_2\}$; clearly $P\in\mathscr{P}$, so $Q_P\subseteq\mathscr{V}$. Let $Q_P=\{V_1,V_2\}$, with $N_1\subseteq V_1$ and $N_2\subseteq V_2$. Then

$$\bigcap\{V\in\mathscr{V}:x\in V\}\subseteq V_1\subseteq X\setminus\{y\}\;,$$

and $y\in X\setminus\{x\}$ was arbitrary, so

$$\bigcap\{V\in\mathscr{V}:x\in V\}=\{x\}\;.$$

Thus, $\mathscr{V}$ is a separating open cover for $X$ of cardinality at most $|\mathscr{N}|$, and it follows that $sw(X)\le nw(X)$.

• Nice. You just beat me to it ... – Henno Brandsma Feb 15 '16 at 21:44