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Let $X$ be the set of all real bounded sequences with the metric $$d(\ (x_n)_{n\in\mathbb{N}}, (y_n)_{n\in\mathbb{N}} \ ) = \sup \{\vert x_n-y_n\vert : n\in \mathbb{N}\}$$ Then, prove that the subset $C$ of all convergent sequences in X is a closed set

I think I have a good way but I'm really having much trouble with manipulating the notation for sequences of sequences. I saw this but the answer was kind of what I've already had in mind (and they don't show the whole proof).

My attempt was to suppose a convergent sequence $(\ (x_n^{(i)})_{n\in\mathbb{N}} \ )_{i\in \mathbb{N}}$ in $C$ and then prove that the limit of $(\ (x_n^{(i)})_{n\in\mathbb{N}} \ )_{i\in \mathbb{N}}$ is a convergent sequence. But I'm really sutcked because I'm not so familiarized with the notation and then I just wrote the definitions and couldn't work with them... so please clarify each step of your answer, if possible.

Marcelo
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1 Answers1

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I have no idea what your $(x_n)^{(i)}_{i\in\mathbb{N}}$ is supposed to mean (what are $i$ and $n$ meaning? Do you mean each $x_n\in C$ and $i$ is the $i$-th component? or the other way round?).

Since we don't need to prove the space $C$ is complete, the proof simplifies.

Pick an arbitrary convergent sequence (of sequences) $y_m\in C$, converging to some $x\in X$. (Each $y_m$ is a sequence $(y_{m,n})_{n\in\mathbb{N}}$). We want to show $x\in C$. Since $\mathbb{R}$ is complete, it suffices to show $x=(x_n)_{n\in\mathbb{N}}$ is Cauchy.

So fix an $\varepsilon>0$. Since $y_n\to x$, for some $y_N$ we have $d(y_N,x)<\varepsilon$, so $|y_{N,n}-x_n|<\varepsilon$ for all $n$. But $y_N\in C$ and so is Cauchy, so $|y_{N,m}-y_{N,m'}|<\varepsilon$ for all $m,m'\geq M$, some $M\in\mathbb{N}$. So $$ |x_m-x_{m'}|\leq |x_m-y_{N,m}|+|y_{N,m}-y_{N,m'}|+|y_{N,m'}-x_{m'}|<3\varepsilon $$ and now go back to change $\varepsilon$ to $\varepsilon/3$ if you must.

user10354138
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