# A net $(x_\alpha)$ has an accumulation point $x$ iff some subnet converges to $x$

I'm looking at the answer to this old question Accumulation point in topological space problem

I figured out the backwards direction on my own, but I'm a little confused about how the subnet is being defined. Here is the excerpt of the proof

"Now suppose that $$x$$ is an accumulation point of $$(x_\alpha)$$. Now define the set $$B= \{(\alpha, U): \alpha \in A, U \text{ open and such that } x_\alpha \in U \}\text{.}$$ Define a partial order by $$(\alpha_1, U_1) \le (\alpha_2, U_2) \text{ iff } \alpha_1 \le \alpha_2, \text{ and } U_2 \subseteq U_1$$. One easily checks that this makes $$B$$ a directed set, and defining $$h: B \rightarrow A$$ by $$h((\alpha, U)) = \alpha$$, and again a small check shows that setting $$y_\beta = y_{(\alpha,U)} = x_\alpha$$ defines a subnet via $$h$$ (do these checks yourself!). And $$(y_\beta)_{\beta \in B}$$ converges to $$x$$: let $$O$$ be an open neighbourhood of $$x$$. Then the set $$A_O$$ (as defined above) is cofinal in $$A$$, so pick any $$\alpha_0 \in A$$ such that $$x_{\alpha_0} \in O$$. Then $$\beta_0 = (\alpha_0, O) \in B$$, and if $$\beta = (\alpha, U) \ge \beta_0$$, we know in particular that $$U \subset O$$, and $$y_\beta = x_\alpha \in U \subset O$$, so indeed all net elements with index larger than $$\beta_0$$ are in $$O$$, as required."

I understand why that is a directed set (take $$(max\{\alpha_1, \alpha_2\}, U_1 \cup U_2))$$). Now I am having trouble understanding what exactly the subnet they are defining is. Is it $$f(h((\alpha, U))) = f(h(\alpha)) = x_\alpha$$? I think its that $$y_\beta$$ throwing me off.

• Take an upper bound for $\alpha_1, \alpha_2$ in $\Lambda$ (by directedness) and $U_1 \cap U_2$, as the order in the second component is reversed, to see directedness. – Henno Brandsma Apr 23 '19 at 21:04
• Moreover I took the index set of the original set to be $A$, instead of $\Lambda$. I find $\alpha \in A$ more logical.... – Henno Brandsma Apr 23 '19 at 21:05

$$(y_\beta)_{\beta\in B}$$ is the new net, so $$y_\beta$$ is defined as $$x_{h(\beta)}$$, that is $$x_\alpha$$ if $$\beta = (\alpha, U)$$.

A net from $$B$$ is defined as a function from $$B$$ to $$X$$ and it’s given by sending $$(\alpha, U)$$ to $$x_\alpha$$, which is the image of $$\alpha$$ under the original net $$A: \to X$$. It’s clear that this is the same as doing $$h$$ (the projection) first and then applying the original net-map and that is what makes $$h$$ the connecting map between these nets. The subnet is denoted $$y_\beta$$ (which is the image of the element $$\beta \in B$$ and not $$x_\beta$$ to avoid confusion: if we call the original net $$x: \Lambda \to X$$ then the subnet can be denoted $$y: B \to X$$ and we have the condition that $$x \circ h = y$$ as maps.

Firstly I think they made a typo, looks like the set $$B$$ should be $$B=\{ (\alpha,U) \,\, : \,\,\alpha \in\Lambda,U \mbox{ open s.t. } x_\alpha\in U \}$$

Secondly, let's review the definitions of net and subnet. A net is a function on a directed set $$f:\Lambda\to X$$, where $$f(\alpha ) \in X$$ is denoted $$x_\alpha$$. A subnet is a co-final increasing function on a directed set $$h:B\to \Lambda$$, where $$h(\beta ) \in \Lambda$$ is denoted $$\alpha_{\beta}$$. Thus given the net $$(x_\alpha)_{\alpha\in\Lambda}$$, the subnet defined by $$h$$ is $$(x_{h(\beta)})_{\beta\in B}=(x_{\alpha_\beta})_{\beta \in B}$$

Then to answer your question, the subnet they are defining is $$h:B\to\Lambda$$ where the elements of $$B$$ are of the form $$\beta = (\alpha , U)$$, and $$h(\beta)= \alpha$$ So the subnet is $$(x_{h(\beta)})_{\beta\in B}$$

I hope this addresses the concern