# A net in $\mathbb{R}$

Let $\{x_j\}_{j\in J}\subset \mathbb{R}$ be a net, $J$ is a directed set.

If $\{x_j\}_{j\in J}$ does not converge to 0, then there is a subnet$\{x_b\}_{b\in B}$, $B$ is a directed set, that $x_b\rightarrow x,$ where $x$ is either $\infty,-\infty,$ or a nonzero real number.

I see this in a proof of the theorem below:

Any finite dimensional topological vector space has the usual Euclidean topology.

I can only prove that there is a convergent subnet, but not surely has nonzero limit, by using the statement below:

A net $\{x_j\}_{j\in J}$ has a cluster point y if and only if it has a subnet $\{y_\beta\}$ that converges to y.

How to prove there is a convergent subnet with nonzero limit?

Any help would be appreciated.

• You don't need to use nets to show that all norms on a finite dimensional space are equivalent. – Henricus V. Mar 31 '16 at 5:20
• @HenryW.yes,but i mean finite dimensional topological space, not normed space. – David Lee Mar 31 '16 at 6:26
• This may be helpful: math.stackexchange.com/questions/445547/… – Henricus V. Mar 31 '16 at 13:51

I will show it for the general case. Let $(X,\tau)$ be a topological space and $\langle x_j \rangle_{j \in J}$ be a net. A point $x \in X$ is a cluster point of $\langle x_j \rangle_{j \in J}$ if and only if there exists a subnet converging to $x$.
The if direction is easy by the definition of directed set. Conversely, suppose that $x$ is a cluster point of $\langle x_j \rangle_{j \in J}$. Define $\mathcal{N}$ to be the neighbourhoods of $x$. Define the preorder $\lesssim$ on $\mathcal{N} \times J$ to be ($U_0,U_1 \in \mathcal{N},j_0,j_1 \in J$)
$$(U_0,j_0) \lesssim (U_1,j_1) \iff U_0 \supseteq U_1 \land j_0 \lesssim j_1$$ For each $(U_0,j_0) \in \mathcal{N} \times J$, we can choose $j_{U_0,j_0} \in J\mid j_{U_0,j_0} \gtrsim j_0, x_{j_{U_0,j_0}} \in U_0$. Then if $(U_1,j_1) \gtrsim (U_0,j_0)$, we have $j_{U_1,j_1} \gtrsim j_1 \gtrsim j_0, x_{j_{U_1,j_1}} \in U_1 \subseteq U_0$. Hence the subnet $$\langle x_{j_{U_0,j_0}} \rangle_{(U_0,j_0) \in \mathcal{N} \times J}$$ converges to $x$.
Classical fact: if every subnet (in any space $X$) of a net has itself a subnet that converges to some fixed $p$, then the original net converges to $p$. (This can be proved using the cluster point fact, if you like.)
Suppose we have net $(x_j)_{j \in J}$ that does not converge to $0$. As $[-\infty, +\infty]$ (the two point compactification of the reals) is compact, there is some subnet that converges to some $p \in [-\infty, +\infty]$ (and hence so do all of its subnets). By the first classical fact, we can in fact choose such a subnet for some $p \neq 0$, or else the net would have converged to $0$ in the first place. QED.