Prove that $nw(X,T) \le w(X,T)$, network weight $\lt$ weight of topological space. Prove that the network weight $nw(X,T)$ is always less than or equal to the weight $w(X,T)$ of a topological space.  I understand how the network weight can equal the weight if you let the network $=$ the basis.  But I can't think of a situation where the network weight would ever be less than the weight of a topological function.
 A: Let $p$ be a free ultrafilter on $\Bbb N$, and let $X=\{p\}\cup\Bbb N$. Let
$$\tau=\wp(\Bbb N)\cup\big\{\{p\}\cup U:U\in p\big\}\;;$$
$\tau$ is a topology on $X$ in which each point of $\Bbb N$ is isolated, and $U\subseteq X$ is an open nbhd of $p$ if and only if $p\in U$ and $U\cap\Bbb N\in p$.
$\langle X,\tau\rangle$ is not first countable at $p$, so $w(X)>\omega$, but $\big\{\{x\}:x\in X\big\}$ is a countable network for $X$.
More generally, any countable space that is not first countable is an example.
Added: A possibly easier example is obtained by letting $p$ be any point not in $\Bbb N\times\Bbb N$, setting $X=\{p\}\cup(\Bbb N\times\Bbb N)$, making each point of $\Bbb N\times\Bbb N$ isolated, and giving $p$ a local base of open nbhds as follows. For each function $f:\Bbb N\to\Bbb N$ and $n\in\Bbb N$ let
$$B(f,n)=\{p\}\cup\{\langle k,\ell\rangle\in\Bbb N\times\Bbb N:k\ge n\text{ and }\ell\ge f(k)\}\;;$$
the set of all such $B(f,n)$ is a local base at $p$.
Clearly $X$ is countable, so it only remains to show that there is no countable local base at $p$. Let $\mathscr{B}=\{U_i:i\in\Bbb N\}$ be countable family of open nbhds of $p$. For each $i\in\Bbb N$ there is a basic open set $B(f_i,n_i)\subseteq U_i$. Now define
$$g:\Bbb N\to\Bbb N:k\mapsto\max_{i\le k}\big(f_i(k)+1\big)\;;$$
clearly $g(k)>f_i(k)$ for each $i\le k$. Let $i\in\Bbb N$ be arbitrary; then
$$\langle n_i,f_i(n_i)\rangle\in B(f_i,n_i)\setminus B(g,0)\subseteq U_i\setminus B(g,0)\;,$$
so $B(f_i,n_i)\nsubseteq U_i$, and $\mathscr{U}$ is not a local base at $p$.
