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In proving the equivalent definitions of compact sets, I come across below equivalence of cluster point. Could you have a check on my proof?

Let $X$ be a topological space, $(x_d)_{d \in D}$ is a net in $X$, and $x \in X$. Let $\mathcal N_x$ be the set of all neighborhoods of $x$. Then the following statement is equivalent:

  • (S1) For each $(U, d) \in \mathcal N_x \times D$, there is $d' \in D$ such that $d' \ge d$ and $x_{d'} \in U$.
  • (S2) There is a sub-net $(x_{\varphi (s)})_{s \in S}$ of $(x_d)_{d \in D}$ such that $x_{\varphi (s)} \to x$.

My attempt:

  • S1 $\implies$ S2

Let $S := \{(U, d) \in \mathcal N_x \times D \mid x_d \in U\}$. We define a pre-order on $S$ by $$(U, d) \le (U', d') \iff U' \subseteq U \text{ and } d \le d'.$$

Let $(U, d), (U', d') \in S$ and $U^* := U \cap U'$. Then $U^* \in \mathcal N_x$. Also, there is $d'' \in D$ such that $d\le d''$ and $d' \le d''$. By S1, there is $d^*\in D$ such that $d^* \ge d''$ and $x_{d^*} \in U^*$. It follows that $(U, d) \le (U^*, d^*)$ and $(U', d') \le (U^*, d^*)$. Then $S$ together with $\le$ is a directed set.

We define a map $\varphi:S \to D$ by $\varphi (U, d) := d$. Clearly, $\varphi$ is increasing. We have $(X, d) \in S$ for all $d \in D$, so $\varphi$ cofinal in $D$. Hence $(x_{\varphi(s)})_{s\in S}$ is a subnet of $(x_d)_{d \in D}$.

Let $U \in \mathcal N_x$. Then there is $d \in D$ such that $x_d \in U$ and thus $(U, d) \in S$. If $(U, d) \le (U', d')$, then $U' \subseteq U$ and thus $x_{\varphi (U', d')}\in U' \subseteq U$. This means $x_{\varphi(s)} \to x$.

  • S2 $\implies$ S1

Assume there is a sub-net $(x_{\varphi (s)})_{s \in S}$ of $(x_d)_{d \in D}$ such that $x_{\varphi (s)} \to x$. Let $(U, d) \in \mathcal N_x \times D$. Then there is $s \in S$ such that $x_{\varphi (s')} \in U$ for all $s' \ge s$. Because $\varphi$ cofinal in $D$, there is $s'' \in S$ such that $\varphi(s'') \ge d$. There is $s^* \in S$ such that $s^*\ge s$ and $s^* \ge s''$. This implies $d^* :=\varphi(s^*) \ge d$ and $x_{d^*} \in U$.

Akira
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    Looks fine. Now you do only use an upperbound... ( see your other compactness proof question and my response) – Henno Brandsma Jan 05 '22 at 01:08
  • @HennoBrandsma Thank you so much for answering many of my questions recently. I really appreciate your kindness. – Akira Jan 05 '22 at 08:27
  • I have nets as a subject I follow. I often use them in proofs. You’re welcome. You seem to be learning about nets now? From what textbook? – Henno Brandsma Jan 05 '22 at 08:29
  • @HennoBrandsma I learn net mainly from Wikipedia. It's unfortunate that we do not have a direct proof using net that sequential compactness implies compactness in metric spaces, i.e., if any sequence has a convergent subsequence then any net has a convergent net. – Akira Jan 05 '22 at 08:37
  • @HennoBrandsma In metric space, any convergent net has a convergent subsequence. So we can easily replace net with sequence in these results ((1) and (2)). – Akira Jan 05 '22 at 08:43
  • that's because a subnet of a sequence need not be a subsequence. It's a common misconception. A subsequence is an example of a subnet but the relation is more subtle. In a first countable space (like a metric space) we can for a cluster point of a net find a subnet based off $\Bbb N$ and with that lemma we're done (I think for compact implies sequential compact in metric spaces). The reverse is more subtle. Think about it. – Henno Brandsma Jan 05 '22 at 08:47
  • @HennoBrandsma I formulate your lemma as follows: Let $X$ be a metric space and $(x_d){d \in D}$ a net in $X$. Then $a \in X$ is a cluster point of $(x_d){d \in D}$ if and only if there is a subnet $(x_{\phi (n)})_{n\in \mathbb N}$ that converges to $a$. – Akira Jan 05 '22 at 09:18
  • @HennoBrandsma Assume the contrary that $(x_d){d \in D}$ is a net in $X$ and has no cluster point. By our lemma, any subnet of the form $(x{\phi (n)}){n\in \mathbb N}$ is not convergent. Let $(x{\phi_0 (n)}){n\in \mathbb N}$ be such a non-convergent subnet. Let $y_n := \phi_0 (n)$. On the other hand, $(y_n){n\in \mathbb N}$ is a sequence and admits a convergent subsequence $(y_{\phi_1(n)}){n\in \mathbb N}$. Clearly, $(y{\phi_1(n)}){n\in \mathbb N}$ is a convergent subnet of $(x_d){d \in D}$, which is a contradiction. – Akira Jan 05 '22 at 09:20
  • @HennoBrandsma Can we show that such $(x_{\phi_0 (n)})_{n\in \mathbb N}$ exists? – Akira Jan 05 '22 at 09:21
  • I guess the answer is negative because you said "A subsequence of a sequence is a subnet as well, but a subnet need not have a cofinal subsequence" here. – Akira Jan 05 '22 at 09:28
  • Let's halt the discussion here. Pose a new question if you want to discuss proofs in metric spaces of this equivalence using nets. The actual question is fine. – Henno Brandsma Jan 05 '22 at 09:37

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