# Limit of sequences in metric space

I could not prove the following statement. Can you help me?

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

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That’s usually taken as the definition of convergence. If you’re supposed to prove it, you must be using some other definition, and you’ll have to tell us what it is. – Brian M. Scott Nov 18 '12 at 0:37
I don’t understand what you’re asking. – Brian M. Scott Nov 18 '12 at 0:43
Title modified. – Did Nov 18 '12 at 9:55

• $(x_n)$ converges to $a$ if for every $\varepsilon >0$ there is a $N$ such that $$n\ge N \Rightarrow \text{d}(x_n, a) <\varepsilon,$$ where $\text{d}$ is the metric function associated to the metric space in question.
• By "$B(a,\varepsilon)$" I mean the ball of radius $\varepsilon$ centered at $a$, which is simply $$\lbrace x \text{ in your metric space}: \text{d}(x,a)<\varepsilon \rbrace$$

$(\Rightarrow)$

Suppose $(x_n)$ converges to $a$. Then for every $\varepsilon>0,$ there is a $N$ for which $$n\ge N \Rightarrow \text{d}(x_n, a)<\varepsilon.$$ Let $U$ be any open neighborhood containing $a$. Since $U$ is open, there is an $\varepsilon >0$ for which $B(a,\varepsilon)\subset U$. Therefore, $n\ge N$ impiles $x_n\in B(a,\varepsilon)\subset U$.

$(\Leftarrow)$

By hypothesis, every open set $B(a, \frac{1}{n})$ contains the tail of some sequence $(x_n)$. Let $\varepsilon >0$. There is an $N$ for which $$n\ge N \Rightarrow \frac{1}{n}<\varepsilon.$$ Therefore, $n\ge N$ implies that $\text{d}(x_n,a)\le \frac{1}{n}<\varepsilon$, since $x_n \in B(a, \frac{1}{n})$ for $n\ge N.$ Therefore, $x_n \rightarrow a$.

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I assume the other definition is $d(x_n,a)\to 0$. If this is to be used, then

Hints: Use the (similar) definition of limit among real numbers, and use that every open set $U\ni a$ (by definition) contains a sphere (ball) around $a$.

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