Let $c_{0}$ be the space of real-valued sequences $\{x_{n}\}$ which converge to zero, equipped with the metric $d(\{x_{n}\}, \{y_{n}\}) = \sup\{|x_{n} – y_{n}|: n \in \mathbb{N}\}$. Let $e_{k}$ denote the sequence in $c_{0}$ which is identically zero, except for the k-th entry which equals 1.

Prove that the metric space $(c_{0}, d)$ is complete.


Consider ${b_n}$, a sequence in $c_0$, which is Cauchy. Then $\forall \epsilon > 0$ $\exists N$, such that if $m \ge N$ and $n \ge N$, then $d(b_m, b_n) < \epsilon$.

So the sequence ${b_n}$ converges to some sequence b. Since all $b_n$ and $b_m$, except of finitely many first elements, are identically zero, b will consist of zeros, except for finitely many elements at the beginning.

Thus $b \rightarrow 0$, so $b \in c_0$. Thus $c_0$ is complete.

Is this proof correct and formal enough?

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    $\begingroup$ "converge to zero" and "only a finite number of non-zero elements" are not the same. $(1/n)_n$ converges to zero, but has no element zero. $\endgroup$ – LutzL Nov 13 '16 at 18:10
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    $\begingroup$ You are missing too many details to cover in comments. First, you need better notation to distinguish the elements $c_0$ and their values at a particular index (I like $x_n \in c_0$ and $x_n(k)$ to refer to the $k$th element. You need to product a candidate limit sequence and then prove that it is in $c_0$ and then that the sequence converges in the above metric. See math.stackexchange.com/q/1952535/27978. $\endgroup$ – copper.hat Nov 13 '16 at 18:15
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    $\begingroup$ I think your metric should be $d(x, y ) = \sup\{|x_{n} – y_{n}|: n \in \mathbb{N}\}$ where $x = (x_n)$ $\endgroup$ – Tom Collinge Nov 13 '16 at 18:22
  • $\begingroup$ To Tom Collinge: Yes, you are right. I have edited the post. $\endgroup$ – user0347284 Nov 13 '16 at 18:28

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