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I cannot prove this statement, I tried to prove by using the definition of open sets however i feel that it is necessary prove it in two directions since it's an iff statement.

The question is,

Let $X$ be a metric space with metric $d$, and let $x_n$ be a sequence of points in $X$. Prove that $x_n\rightarrow a$ if and only if for every open set $U$ with $a\in U$, there is a number N such that whenever n > N we have $x_n\in U$

What I have done is, by using the definition of open sets, $B(a,r)\cap U$ is not an empty set and $r > 0$. So by choosing $r=1$ we obtain $x_1\in B(a,1)$ which intersects with $U$. Secondly I chose $r=1/2$ so $x_2\in B(a,1/2)$ which again intersects with $U$. By continuing like this eventually I chose $r=1/n$ so $x_n\in B(a,1/n)$ which intersects with $U$. Therefore $d(x_n,a) <\frac{1}{n}$. Therefore I conclude that when $n$ goes to infinity that $x_n\rightarrow a$.

Am I right up to this point? If so how should I proceed and what should I do to prove that for opposite inclusion?

I'm so new, so let me know if I do or did something inconvenient. Thanks!

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Looks fine so far. For the other direction, choose various $U$ wisely. Given $\epsilon>0$ let $U=B(a,\epsilon)$, ..., $d(x_n,a)<\epsilon$ for all $n>N$. – Hagen von Eitzen Nov 25 '12 at 21:00


The condition $B(a,r)\cap U\ne\varnothing$ is used to show that $a$ is a limit point of $U$. To show that $U$ is open, $$ \forall a\in U,\exists r>0:B(a,r)\subset U $$


The standard metric space concept of convergence is $$ \forall\epsilon>0,\exists N:\forall n\ge N,d(x_n,a)<\epsilon $$

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I'm not sure why you've done what you've done, or what it proves.

To prove the if part of your theorem, suppose we have a sequence $x_n$. For any $\epsilon > 0$, there exists $N$ such that $x_n\in B(a,\epsilon)$ $\forall n > N $ (since $B(a,\epsilon)$ is an open set containing $a$) . Hence $x_n \rightarrow a$.

For the only if part, suppose $x_n \rightarrow a$. Let $U$ be an open set containing $a$. Then by the definition $\exists r >0$ such that $B(a,r) \subset U $. Then $\exists N$ such that $x_n \in B(a,r) \forall n >N$ since $x_n \rightarrow a$. Hence $x_n \in U$ $ \forall n > N$

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