I'm studying the proof that space $c$ is complete.
The space c consists of all convergent sequences $x=(\mathcal{E}_j)$ of complex numbers, with the metric induced from the space $l^\infty$.
Proof. $c$ is a subspace of $l^\infty$ and we show that $c$ is closed in $l^\infty$.
We consider any $x=(\mathcal{E}_j)\in\bar c$, closure of $c$. Hence, there are $x_n=(\mathcal{E}_j^{(n)})\in c$ such that $x_n \rightarrow x$. Hence, given any $\epsilon>0$, there is an $N$ such that for $n \ge N$ and all $j$ we have $$\vert \mathcal{E}_j^{(n)} - \mathcal{E}_j \vert \le d(x_n,x) \lt \frac \epsilon 3, $$ in particular, for $n=N$ and all $j$.
I have the following questions, wonder if someone could help me with them.
Questions:
1) An example of $x=(\mathcal{E}_j)\in\bar c$. For simplicity, I think of a sequence $\{1,1,...,1\}$. I don't know with certainty if this is correct.
2) An example of $x_n=(\mathcal{E}_j^{(n)})\in c$ such that $x_n \rightarrow x$. If $x \in \bar c = \{1,1,...,1\}$. Then, what is an example of $x_n$ such that $x_n \rightarrow x$?
Thank you!
Isn't $(1,1,1,\dots)$ a convergent sequence and thus is in $c$?
To show $c$ is closed in $l^\infty$, suppose $x^{(k)} \to x$ where $x^{(k)} = (x_n^{(k)})_n$ and $x = (x_n)_n$ are sequences. Let $L_k = \lim_n x_n^{(k)}$. Then, since $(x^{(k)})_k$ is Cauchy in $l^\infty$, it's easy to see that $(L_k)_k$ is Cauchy and thus converges to some complex number $L$.
We show $x_n \to L$. Take $\epsilon > 0$. Take $k$ large so that $|L-L_k| < \epsilon$ and then $||x-x^{(k)}||_\infty < \epsilon$. Then for $n$ large enough, $|x^{(k)}_n-L_k| < \epsilon$ and so $|x_n-L| \le |x_n-x^{(k)}_n|+|x^{(k)}_n-L_k|+|L_k-L| < 3\epsilon$ for $n$ large enough.