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I am having trouble solving the exercise 4 in chapter 2, section 4 of Introduction to Topology, Gamelin and Greene, 2nd.

Suppose a topological space $X$ satisfies the first axiom of countability, or is first countable, i.e., for each $x\in X$, there exists a sequence of open neighborhoods $\{ U_n\}$ of $x$ such that each neighborhood of $x$ includes one of the $U_n$'s.

Prove the following assertions:

(c) In a first-countable space $X$, any point adherent to a set $S$ is a limit of a sequence in $S$.

Authors suggest that because $S$ meets each $U_n$, just pick any $s_n\in U_n \cap S$ for each n. Then $\{s_n\}$ converges to $x$.

I don't get the last part that "$\{s_n\}$ converges to $x$".

To me it seems reasonable that for any open neighborhood $V$ of $x$, there is some $U_k\subset V$ thus $s_k \in V$. But to say that $\{s_n\}$ converges to $x$, it is necessary to show that there exists some $N$ such that if $n\geq N$ then $s_n \in V$. I have no way to show the existence of such $N$.

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    $\begingroup$ There won't be such an $N$ in general, you have to pick the $s_n$ so that there is. You need something like $s_m \in U_n $ for any $m \geq n$, can you see how to get that? $\endgroup$ – David Hartley Sep 29 '20 at 15:26
  • $\begingroup$ @DavidHartley Oh I think I get it now. Maybe I misunderstood the authors' intention. $\endgroup$ – Henry Choi Sep 29 '20 at 15:29
  • $\begingroup$ @DavidHartley I leave my answer by your hint below though I am not sure I went right way. Thank you. $\endgroup$ – Henry Choi Sep 29 '20 at 15:48
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    $\begingroup$ The suggestion fails since we do not know much about the $U_n$. For example it is possible that infinitely many $U_n = X$ which allows to choose $s_n = y \ne x$ for these $n$. The resulting sequence does not converge to $x$. $\endgroup$ – Paul Frost Sep 29 '20 at 17:15
  • $\begingroup$ @PaulFrost I got a kick out of your counterexample. $\endgroup$ – Henry Choi Sep 30 '20 at 1:18
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If $\{U_n\}_{n\in\omega}$ is an open local base for a point $x$ you always can suppose WLOG that $U_{n+1}\subseteq U_n$ (You only define $V_n:=\cap_{i\leq n}U_n$, then $\{V_n\}_{n\in\omega}$ is also an open local base).

Note that under this assumption you get $s_n\in U_N$ for every $n\geq N$.

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  • $\begingroup$ I did not catch I could make the chain of open sets when I posted this question. I think my answer goes to the same direction as your answer. Thank you. $\endgroup$ – Henry Choi Sep 29 '20 at 16:02
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With Hartley's comment I could go this way.

Fix an arbitrary $s_1 \in U_1 \cap S$.

Having picked $s_1, s_2, \dots, s_{n-1}$, choose $s_n \in (\cap_{k=1}^{n} U_k)\cap S$. Because $\cap_{k=1}^{n} U_k$ is an open neighborhood of $x$ there exists some $U_\alpha \subset \cap_{k=1}^{n} U_k$. Pick any $s_n \in U_\alpha \cap S$. Then $s_n \in (\cap_{k=1}^{n} U_k)\cap S$.

Fix a neighborhood $V$ of $x$. Then by the definition of the first axiom of countability, there exists $U_m \subset V$ for some $m$. If $n\geq m$, $s_n \in S_m \subset V$.

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    $\begingroup$ Slightly simpler approach: Define $V_n = \bigcap_{k=1}^n U_k$ and pick $s_ n \in V_n \cap S$. Let $V$ be a neighborhood of $x$. It contains some $U_k$. But then for $n \ge k$ we have $s_n \in V_n \subset U_k \subset V$. $\endgroup$ – Paul Frost Sep 29 '20 at 17:23
  • $\begingroup$ I have to admit this is more consistent with the another answer above. $\endgroup$ – Henry Choi Sep 30 '20 at 1:11

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