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My attempt:

$\overline{B} \subseteq \ell^\infty$ follows from the fact that $B \subseteq \ell^\infty$ and $\ell^\infty$ is closed.

For $\ell^\infty \subseteq \overline{B}$, we must show every sequence in $\ell^\infty$ can be represented as a limit point of a sequence in $B$. Thus, consider $x = (x_1, x_2, ...) \in \ell^\infty$. If $x$ has a constant subsequence, we are done, so consider when $x$ does not. Let $d_i = (x_1, x_2, ..., x_i, \sup\{x_{i+1}, x_{i+2}, ...\}, \sup\{x_{i+1}, x_{i+2}, ...\}, ...\}$ . Thus, $d_i$ is an element of $B$ as it has a constant subsequence.

Now, consider the case where $x$ does not reach its supremum on any of its elements. Thus, $x$ must have an infinitely increasing or decreasing tail whose elements converge to the supremum. Thus, by nature of the supremum, we can find a $N \in \mathbb{N}$ such that when $n > N$, $||x - d_n|| < \epsilon$. Thus, we need only consider when $x$ has no infinitely decreasing/increasing tails, so reaches its supremum on all tails. Formally, $||x||_\infty$ is reached by some $x_i$ on each tail of $x$. Thus, since $x$ has no constant subsequence, $||x||_\infty$ can only be repeated a finite number of times in the sequence. Thus, we can form a decreasing chain of maximum elements in $||x||_\infty$. Thus, $||x - d_i||$ form an eventually decreasing sequence as $i \rightarrow \infty$. However, at this point I am having trouble proving that $||x - d_i||$ becomes $\epsilon-$close to each other and can only really state that they form a decreasing chain.

For the second question, i.e. showing what $int(B)$ is, I believe $int(B) = \emptyset$. Suppose for a contradiction $B_r(x) \subseteq B$ for some $x \in B, r > 0$. However, consider the sequence $y$ which replaces each constant subsequence $x_{n_k}$ of $x$ with the subsequence $x_{n_k} + r/2^k$ and is the same otherwise. Thus, $||x - y|| = r/2$ so $y \in B_r(x)$ but $y \not \in B$, so this open ball cannot be contained in $B$. Contradiction.

At this point, I would really appreciate any feedback on whether or not I am on the right track for both of these proofs (and if the latter looks complete or not). Thank you!

learning_linalg
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    $B_r(x)\subset B\subset \ell^{\infty}$ implies $x\in \ell^{\infty}$ lateron You treat is as a number. Or do you mean the value of the constant subsequence? – Peter Melech Aug 09 '20 at 14:44
  • Now it makes more sense, don´t You actually mean $y=(||x||{\infty}+\frac{r}{2^i}){i=1}^{\infty}$ where $||x||{\infty}=\sup{i\in\mathbb{N}}|x_i|$. It is not clear to me why $y\in B_r(x)$, i.e. why $||x-y||{\infty}=||(x_i-||x||{\infty}-\frac{r}{2^i}){i=1}^{\infty}||{\infty}<r$? – Peter Melech Aug 09 '20 at 15:12
  • Well, now I am not sure how You know $y\notin B$ since $x$ can have infinitely many constant subsequences , and so it is not clear why $y$ as constructed has no constant susequence... – Peter Melech Aug 09 '20 at 15:23
  • OK, I still have some doubts that might not be justified though, but isn´t it possible that if $x_1,x_2,...$ are the disinct values of the constant subsequences that $x_1+\frac{r}{2}=x_2+\frac{r}{4}=x_3+\frac{r}{8}=...$ etc? – Peter Melech Aug 09 '20 at 15:36
  • I don´t see how that would contradict the countability, and I thought that Chrystomath is right that it can be modified so that it has no constant subsequence, but now I have doubts about that, too – Peter Melech Aug 09 '20 at 16:12

1 Answers1

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$\overline{B}=\ell^\infty$.

Proof: Let $x=(x_1,x_2,\ldots)\in\ell^\infty$ and let $\epsilon>0$. Then the set of values $\{x_n:n\in\mathbb{N}\}$ is bounded by $\|x\|_\infty$, hence has a convergent subsequence. Pick any of these limit points, but let's pick $\alpha=\limsup x$ as OP did. Then by definition there is a sequence of indices $n_i$ such that $$i>N\implies|x_{n_i}-\alpha|<\epsilon$$

Construct the sequence $y:=(y_n)$ such that $y_n=\begin{cases}\alpha&\exists i, n=n_i,\\x_n&\forall i, n\ne n_i\end{cases}$. Since $n=n_i$ for an infinite number of times, $y$ has a constant subsequence and so $y\in B$. Also $$\|x-y\|_\infty=\sup_n|x_n-y_n|=\sup_i|x_{n_i}-\alpha|<\epsilon$$ so $y\in B_\epsilon(x)$, and $B$ is dense in $\ell^\infty$.

The interior of $B$ is empty since any sequence $y$ with a constant sub-sequence can be modified by small amounts that destroy the constant sub-sequence. For example, if $y$ has a single constant sub-sequence, then $y+(1/n)_{n>N}$ is within $1/N$ of $y$ but has no constant sub-sequence.

Chrystomath
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    +1 for the first part, the second is a little bit vague, how do you proceed if the considered sequence has infintely many constant subsequences? – Peter Melech Aug 09 '20 at 15:29
  • @PeterMelech Agreed, it's vague. There will be a countable number of constant subsequences, so one can modify each. It's just messy to write up, without clarifying the concept. – Chrystomath Aug 09 '20 at 15:32
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    I agree! It is messy to write up, but I understand what you mean – Peter Melech Aug 09 '20 at 15:38
  • Yes it applies to $\mathbb{C}$ which is finite-dimensional, a bounded sequence of sequences might have no convergent subsequence but that´s something different – Peter Melech Aug 09 '20 at 15:57