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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.

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  • $\begingroup$ 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. $\endgroup$ – Henno Brandsma Apr 23 '19 at 21:04
  • $\begingroup$ Moreover I took the index set of the original set to be $A$, instead of $\Lambda$. I find $\alpha \in A$ more logical.... $\endgroup$ – Henno Brandsma Apr 23 '19 at 21:05
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$(y_\beta)_{\beta\in B}$ is the new net, so $y_\beta$ is defined as $x_{h(\beta)}$, that is $x_\alpha$ if $\beta = (\alpha, U)$.

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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.

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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

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