# A property of a first-countable space

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 $$N$$.

• 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? – David Hartley Sep 29 '20 at 15:26
• @DavidHartley Oh I think I get it now. Maybe I misunderstood the authors' intention. – Henry Choi Sep 29 '20 at 15:29
• @DavidHartley I leave my answer by your hint below though I am not sure I went right way. Thank you. – Henry Choi Sep 29 '20 at 15:48
• 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 $x$. – Paul Frost Sep 29 '20 at 17:15
• @PaulFrost I got a kick out of your counterexample. – Henry Choi Sep 30 '20 at 1:18

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 $$n\geq N$$.

• 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. – Henry Choi Sep 29 '20 at 16:02

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 $$n\geq m$$, $$s_n \in S_m \subset V$$.

• 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 $s_n \in V_n \subset U_k \subset V$. – Paul Frost Sep 29 '20 at 17:23
• I have to admit this is more consistent with the another answer above. – Henry Choi Sep 30 '20 at 1:11