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Let $(M,d)$ be a metric space, let ${x_n}\in M$ and $p\in M$

Prove: $x_n\to p \iff d(x_n,p)\to 0$

$\Leftarrow:$ be definition of a limit, for all $0 < \varepsilon$ there is $N\leq n$ such that $$|d(x_n,p) + 0|< \varepsilon \iff |d(x_n,p)|< \varepsilon \iff d(x_n,p)< \varepsilon$$

last $\iff$ is due to $d:M\times M\to [0,\infty)$

$\Rightarrow:$ $x_n\to p$, let $\varepsilon = \frac{1}{n}$ therefore for all $N\leq n$ $$0 \leq d(x_n,p)<\frac{1}{n}$$ using the squeeze theorem we get $$d(x_n,p)\to 0$$

Have I missed something? is it correct?

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newhere
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1 Answers1

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The ideas are all there, but your argument is somewhat muddled. Note the ordering, especially when it comes to $N$ and $n$:

If $d(x_n,p) \to 0$ then for any $\epsilon > 0$ there exists a natural number $N$ such that for any $n \geq N$ we have $$d(x_n,p) < \epsilon.$$ (Absolute values not necessary as distances are non-negative.) But this is precisely the condition for $x_n \to p$ in a metric space, so the two statements are equivalent.

DanLewis3264
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