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Let $M$ be a metric space, $A\subseteq M$. $A$ is closed in $M$ $\iff$ for all sequences $\{x_{n}\}$ in $A$ which converge in $M$, the limit points are in $A$.

I get the proof of $\Rightarrow$

In the lecture note the proof for $\Leftarrow$ is written as follow:

Let assume that $A$ is not closed, and let us build a sequence of points in $A$ that converges to a point outside of $A$. $A$ is not closed, therefore $A^c$ is not open, so there is $p\in A^c$ such that, for all $\epsilon>0$, $B(p,\epsilon)\not\subseteq A^c$ in particular it is true for $\epsilon=\frac{1}{n}$ which means that for all $n$ there are points $x_n\in A$ but $x_{n}\in B(p,\frac{1}{n})$ therefore $d(x_n,p)<\frac{1}{n}$ and $x_{n}\rightarrow p$ and $x_n\in A$ and for all $n$ $p\in A^c$

I did not understand that move "which means that for all $n$ there are points $x_n\in A$ but $x_{n}\in B(p,\frac{1}{n})$ therefore $d(x_n,p)<\frac{1}{n}$ and $x_{n}\rightarrow p$ and $x_n\in A$ and for all $n$ $p\in A^c$" what have he shown here?

egreg
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gbox
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  • Limits of sequences are unique in metric spaces so we can say: for all sequences in $A$ that converge to $x \in M$, $x \in A$ as well. "the limit points" is confusing; there is one unqie limit for a convergent sequence. – Henno Brandsma Aug 07 '17 at 21:53

2 Answers2

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Statement A: "every sequence in $A$ that converges, converges in $A$ $\implies A$ is closed"

is equivalent to Statement B: "$A$ is not closed $\implies$ there is converging sequence in $A$ that converges to a point not in $A$".

So to prove that, he assumes $A$ is not closed, and then he mechanically creates a $\{x_n\}$ so that $x_n \rightarrow p \not \in A$. By creating that sequence, he will have proven statement B. And therefore he will have proven statement A. (which is half of the proof).

$A$ not closed $\implies$ $A^c$ not open $\implies$ there is a point $p \in A^c$ so that $p$ is not an interior point of $A^c$ $\implies$ every neighborhood $N_{\frac 1n}(p)$ will have a point $x_n \in N_{\frac 1n}(p)$, $x_n \in (A^c)^c = A$ $\implies $ $\{x_n\}\subset A$, $x_n \rightarrow p \not \in A$.

So statement B has been proven.

fleablood
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By showing this you have got a contradiction to your hypothesis: "for all sequences $\{x_{n}\}$ in $A$ which converge in $M$, the limit points are in $A."$

You have created a sequence in $A$ which coverges to a point outside $A$.

Sahiba Arora
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