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A point $c\in X$ is a cluster point of the net $(x_d)_{d\in D}$ if, for every neighborhood $U$ of $c$ and for any $d_0\in D$ there exists $d\ge d_0$ such that $x_d\in U$. In the other words, $x_d$ is frequently (cofinally) in $U$.

Question: How to show that for any cluster point $c$ of $(x_d)_{d\in D}$ there is a subnet converging to $c$?

Since this result is often used in connection with nets, I considered useful to have the proof available somewhere on the site.

It is worth mentioning that different definitions of subnet are commonly used: Different definitions of subnet. (Although for our purpose they are similar in the sense that they give the same set of limits of convergent subnets.)

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Let us define the directed set $$D'=\{(U,d); U\text{ is a neighborhood of }c, d\in D, x_d\in U\}$$ with the natural ordering, i.e., $$(U,d)\le (U',d') \Leftrightarrow (U\supseteq U') \land (d\le d').$$ This is indeed a directed set, if $x_{d_1}\in U_1$ and $x_{d_2}\in U_2$ then there is $d\in D$ such that $d\ge d_1$, $d\ge d_2$ and $x_d\in U_1\cap U_2$.

Let us define $f\colon D'\to D$ by $f(U,d)=d$. The map $f$ is clearly cofinal and monotone. We see that $x_{(U,d)}=x_d$ and that $(x_{(U,d)})$ is a subnet of $(x_d)$.


Two or three various notions of subnet are commonly used.1 But we have found a map which is both monotone and cofinal - which is the most restrictive of the three definitions.


1I wrote two or three since AA-subnet is mentioned less frequently, so this definition is probably less common. See, for example: Different definitions of subnet on this site, my notes on various definition of subnets which are based on Schechter's book.

  • Yes, the standard product trick and using reversely ordered neighbourhoods. To see that $x \circ f$ converges to $c$: For $U$ an open neighbourhood of $c$, define $d' = (U,d)$ for any $d \in D$, and if $(V,d') \ge (U,d)$ we have $x_{d' } \in V \subseteq U$ etc. – Henno Brandsma May 03 '20 at 15:39
  • Should the last line before the second paragraph have $x_{d}\in U_{1}\cap U_{2}$ as opposed to $d\in U_{1}\cap U_{2}$ ? – Travis C Cuvelier Jul 13 '23 at 15:56
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    @TravisCCuvelier Thanks for noticing. – Martin Sleziak Jul 14 '23 at 17:19