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I have a set of all bounded sequences of real numbers with the metric $d$ defined as:

d(a, b) = sup$_n|a_n - b_n|$.

I want to prove this is a complete metric space, i.e. every Cauchy sequence is convergent.

My attempt at Solution:

Consider a Cauchy sequence $a_n$. For every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that for all $m, n > N$, we have sup$_{n, m} |a_n - a_m| < \epsilon$. $\implies $ $|a_n - a_m| < \epsilon$ for all $n, m > N$.

Every bounded sequence in reals has a convergent subsequence - call this subsequence $a_{n_k}$. Hence, for every $\epsilon > 0$, there exists an $M \in \mathbb{N}$ such that for all $k > M$, we have $|a_{n_k} - L| < \epsilon$ $\implies $ sup$_k |a_{n_k} - L| < \epsilon$.

Let $N^* = \max \{N, n_M \}$ so that for all $n, m > N^*$ and $n_k > N^*$ for some $k > M^*$, we have:

sup$_k|a_{n_k} - L| < \epsilon$ and

sup$_{n, m}|a_n - a_m | < \epsilon$.

Then,

$|a_n - L | = |a_n - a_{n_k} + a_{n_k} - L| \leq |a_n - a_{n_k}| + |a_{n_k} - L| $.

Now if $k > M^*$ such that $n_k > N^* \geq N$ and also $n > N^*$, we have

$|a_n - L| \leq 2 \epsilon$

Hence, sup$_n |a_n - L| \leq 2 \epsilon$.

rims
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    If the sequence is Cauchy in this metric, then each coordinate of the sequence is uniformly Cauchy in the usual metric on $\Bbb R$. Show the coordinate-wise limit of the Cauchy sequence yields the limit of the Cauchy sequence. – Robert Shore Sep 22 '19 at 00:46
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    Also notice that each member of your cauchy sequence is again a sequence. So taking a cauchy sequence $(a_n)$ means that for every $\varepsilon > 0$ $\dots$ , we have $d(a_n,a_m) < \varepsilon$ where $d(a_n,a_m) = \sup_{k\in\mathbb{N}} |a_n^{(k)} - a_m^{(k)}|$, where the $a_n^{(k)},a_m^{(k)} \in \mathbb{R}$ are the $k$-th components of the $n$ (respectively) $m$-th sequence in the picked Cauchy sequence. – hal4math Sep 22 '19 at 00:55
  • @RobertShore I am not sure I understand your comment; can you take a look at my full solution method above. Does it make sense or not? – rims Sep 22 '19 at 02:07
  • No, I don't think you have written a correct proof. You seem to be mixing up each individual sequence with the sequence of sequences. – Robert Shore Sep 22 '19 at 02:40
  • According to my understanding, the object $a$ is a sequence, $a_n$ is the nth component of the sequence where $a_n \in \mathbb{R}$. Clearly, this is not the case. What's the interpretation of $a_n^{(k)}$? – rims Sep 22 '19 at 10:06

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