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I know that:

Definiton 1. The sequence $(x_n)$ in the metric space $(X,d)$ is said to converge to the point $x_0\in X$ if $$\forall\epsilon>0, \exists n_0\in\mathbb{N} \text{ such that } \forall n\geq n_0, d(x_n,x_0)<\epsilon.$$

In other words, the sequence $(x_n)$ in the metric space $(X,d)$ converges to the point $x_0\in X$ if $d(x_n,x_0)\rightarrow 0$ with $n\rightarrow\infty.$

Definiton 2. The sequence $(x_n)$ in the metric space $(X,d)$ converges to the point $x_0\in X$ if in every neighborhood $U_{x_0}$ of $x_0$ there exists a natural number $n_0$ such that it is satisfied $\forall n\geq n_0\Rightarrow x_n\in U_{x_0}.$

The point $x_0$ is said to be the limit of the sequence $(x_n)$. Write $x_n\rightarrow x_0, (n\rightarrow\infty)$ or $\lim_{n\to\infty}x_n=x_0.$

Can these two definitions be proved to be equivalent?

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  • $\begingroup$ What does "was the replies a natural number n0 such that to be completed" mean? $\endgroup$ – DonAntonio Nov 9 '13 at 20:26
  • $\begingroup$ now corrected, I think now you clearer, thank you $\endgroup$ – Madrit Zhaku Nov 9 '13 at 20:29
  • $\begingroup$ Not really. First, I think your "suburb" is better translated as "neighborhood", and second: instead of all that you wrote, is enough to write "...every neighborhood $\;U_{x_0}\;$ of $\;x_0\;$ there exists $\;n_U\in\Bbb N\;$ s.t. $\;n>n_U\implies x_n\in U_{x_0}\;$ " $\endgroup$ – DonAntonio Nov 9 '13 at 20:31
  • $\begingroup$ oh, ok sir, thanks $\endgroup$ – Madrit Zhaku Nov 9 '13 at 20:32
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Where are you stuck? As some starting food for thought, notice that

1) For every open neighborhood $U_{x_0}$ of $x_0$, there exists an $\epsilon>0$ with $(x_0-\epsilon,x_0+\epsilon)\subset U_{x_0}$.

2) For every $\epsilon>0$, the set $(x_0-\epsilon,x_0+\epsilon)$ is an open neighborhood of $x_0$.

These two facts should suggest that both definitions have the same content.

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Let $x_{n}$ converge to $x_{0}$ according to definition 1) and let $N$ be a neighbourhood of $x_{0}$.

Since we are working in a metric space this means exactly that some $\epsilon>0$ exists such that $d\left(x_{0},y\right)<\epsilon\Rightarrow y\in N$.

Then $n_{0}\in\mathbb{N}$ exists such that $n\geq n_{0}$ implies that $d(x_{0},x_{n})<\epsilon$ and consequently $x_{n}\in N$.

Conversely let $x_{n}$ converge to $x_{0}$ according to definition 2) .

Note that for every $\epsilon>0$ the set $N=\left\{ y\in X\mid d\left(x_{0},y\right)<\epsilon\right\} $ is a neighbourhood of $x_{0}$, hence some $n_{0}\in\mathbb{N}$ exists such that $n\geq n_{0}$ implies that $x_{n}\in N$ or equivalently $d(x_{0},x_{n})<\epsilon$.

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