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I asked a question earlier today, but I still feel like there is room for another question here.

For the following, $X$ is a metric space, $A\subseteq X$, and the notation $\lbrace a_n \rbrace_{n\in \mathbb{N}}\rightarrow x$ denotes the sequence converging to the limit $x\in X$.

$\textbf{(1)}$ $\forall$ sequence $\lbrace a_n \rbrace_{n\in \mathbb{N}}$ that is a subset of $A$, [if $\lbrace a_n \rbrace_{n\in \mathbb{N}}\rightarrow x$, then $x\in A$].

<p>$\textbf{(2)}$ the set $A$ is closed (i.e. $\forall \text{ limit point } c \text{ of } A, c\in A.$).</p>

$\textbf{Proof for why (1) implies (2):}$ Assume $\forall$ sequence $\lbrace a_n \rbrace_{n\in \mathbb{N}}$ that is a subset of $A$, [if $\lbrace a_n \rbrace_{n\in \mathbb{N}}\rightarrow x$, then $x\in A$]. Now, let $c$ be an arbitrary limit point of $A$. Thus, it is enough to show $c\in A$ to complete the proof. From the definition of a limit point, it is true that $\forall r >0, (B_r(c)-\lbrace c\rbrace)\cap A\neq \phi$. Hence, $\forall n\in \mathbb{N}, (B_\frac{1}{n}(c)-\lbrace c\rbrace)\cap A\neq \phi$. Thus, $(\exists \lbrace x_i \rbrace_{i\in \mathbb{N}}\subseteq A)( \forall n\in \mathbb{N}) [x_n\in (B_\frac{1}{n}(c)-\lbrace c\rbrace)]$. Say $\lbrace a_i \rbrace_{i\in \mathbb{N}}$ is such a sequence that satisfies this argument. Thus, by the original assumption, it is enough to show $\lbrace a_i \rbrace_{i\in \mathbb{N}}\rightarrow c$ to figure $c\in A$ to complete the proof.

$\textbf{Question}:$ Why/how does $\lbrace a_i \rbrace_{i\in \mathbb{N}}\rightarrow c$?

joriki
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W. G.
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  • Take $a_n=x_n\in A\cap (B_{1/n}(c)$ \ ${c}).$ The distance from $a_n$ to $ c$ is less than $1/n$ so the sequence $a_n\to c.$.. Now re-write (1) using the letter $ c$ for the letter $x.$ – DanielWainfleet Jul 11 '18 at 01:09

1 Answers1

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When you say that $\{a_i\}_{i\in\mathbb{N}}$ is a sequence that satisfies this argument you are saying that $a_n\in B_{\frac{1}{n}}(c)-\{c\}$, so $d(a_n,c)<\frac{1}{n}$ and convergence follows.

The outline of the proof for $(1)\implies (2)$ is to construct a sequence using the definition of limit point which, by construction, converges to the limit point. Then, the assumption $(1)$ gives that the limit point is contained in the set, and we have $(2)$.

Jürgen Sukumaran
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  • I understand what you are saying, but I am stuck on the convergence part to $c$ after that. So, let $\epsilon >0$. What natural number $N$ do I choose to have this converge to $c$? – W. G. Jul 10 '18 at 22:39
  • Choose $N$ such that $\frac{1}{N} < \epsilon$. This can be done since there is no "biggest' natural number we can always take $N > \frac{1}{\epsilon}$ – Jürgen Sukumaran Jul 10 '18 at 22:40
  • I think I got it. So, let $\epsilon >0$. Say $N$ has the property that $\frac{1}{N}<\epsilon$. Now, let $n\geq N$. Thus, $d(a_n, c) <\frac{1}{n}\leq \frac{1}{N}<\epsilon$. Is that correct? – W. G. Jul 10 '18 at 22:47
  • Never mind. I got it! Nice answer! Thank you – W. G. Jul 10 '18 at 22:53