Bolzano Weierstrass Theorem for Nets? For real sequences, we know that:
(1) Every bounded sequence has a convergent subsequent.
(2) Every sequence has a monotone subsequence.
For nets, do we have the corresponding theorems? (Extra: Can you give me reference?) Thank you.
 A: Theorem (1) is true for nets: any bounded net in $\mathbb{R}^n$ has a convergent subnet.  Indeed, more generally, a topological space $K$ is compact iff every net in $K$ has a convergent subnet.  This is a standard theorem you should be able to find in any text that covers nets.  In fact, you can even find a proof on Wikipedia.
Theorem (2) is also true for nets: any net in $\mathbb{R}$ has a monotone subnet.  To prove this, let $(x_i)$ be a net in $\mathbb{R}$.  Passing to a subnet using Theorem (1), we may assume $(x_i)$ converges to some $x\in\mathbb{R}\cup\{-\infty,\infty\}$.  Further passing to a subnet, we may assume either $x_i<x$ for all $i$, $x_i=x$ for all $i$, or $x_i> x$ for all $i$ (since at least one of these properties must hold for a cofinal set of $i$).  The second case is trivial and the first and third cases are essentially equivalent; without loss of generality, suppose we are in the first case.
Let $(I,\leq)$ be the index set of the net and define a new order $\preceq$ on $I$ by saying $i\preceq j$ iff $i\leq j$ and $x_i\leq x_j$.  I claim the order $\preceq$ is directed.  Indeed, given $i,j\in I$, there exists $k$ such that $i\leq k$ and $j\leq k$.  Since $x_i<x$ and $x_j<x$ and the net converges to $x$, there then exists $\ell\geq k$ such that $x_\ell>\max(x_i,x_j)$.  We then have $i\preceq\ell$ and $j\preceq\ell$.
Thus $(I,\preceq)$ is also a directed set.  The identity map $(I,\preceq)\to(I,\leq)$ is cofinal, and so we get a subnet of $(x_i)$ where the index set is ordered by $\preceq$ instead of $\leq$.  This subnet is monotone increasing, by the definition of $\preceq$.
(This argument works with $\mathbb{R}$ replaced by any totally ordered set with the order topology, except that to find a limit $x$ of a subnet you may need to pass to the Dedekind completion of the totally ordered set.)
