# Prove that the space of divergent sequences in $(l_{\infty},d_{\infty})$ is open and dense. Is it separable?

The problem statements are:

Consider the space $A=\{ \{a_n\}_{n \in \mathbb N} \in l_{\infty} : \{a_n\}_{n \in \mathbb N} \text { is not convergent }\}$

$a)$ Prove that $A$ is open and dense in $(l_{\infty},d_{\infty})$

$b)$ Decide if $A$ is a separable metric subspace of $(l_{\infty},d_{\infty})$.

My attempt at a solution:

For $a)$, I found it easy to prove density:

Take $\{a_n\}_{n \in \mathbb N}$ and let $\epsilon>0$, if $\{a_n\}_{n \in \mathbb N} \in A$, then it's immediate that $B(\{a_n\}_{n \in \mathbb N}, \epsilon) \cap A \neq \emptyset$. Now, suppose $\{a_n\}_{n \in \mathbb N} \in A^c$, i.e., $\{a_n\}_{n \in \mathbb N}$ is a convergent sequence. Define $\{y_n\}_{n \in \mathbb N}$ as $y_1=a_1 + \dfrac{\epsilon}{2}$, $y_n=a_n+\dfrac{\epsilon}{4}$ if $n$ is even and $y_n=x_n+\dfrac{\epsilon}{3}$ if $n$ is odd and $n>1$.

We have that $|a_n-y_n|\leq |a_1-y_1|=\dfrac{\epsilon}{2}$, which means $d_{\infty}(\{a_n\}_{n \in \mathbb N},\{y_n\}_{n \in \mathbb N})=\sup_{n \in \mathbb N} |a_n-y_n|=\dfrac{\epsilon}{2}<\epsilon$.

The sequence $\{y_n\}_{n \in \mathbb N}$ can be expressed as $\{y_n\}_{n \in \mathbb N}=\{a_n\}_{n \in \mathbb N}+\{b_n\}_{n \in \mathbb N}$, where $\{b_n\}_{n \in \mathbb N}$ is defined as $b_1=\dfrac{\epsilon}{2}$, $b_n=\dfrac{\epsilon}{4}$ if $n$ is even, and $b_n=\dfrac{\epsilon}{3}$ if $n$ is odd and $n>1$. We want to show that $\{y_n\}_{n \in \mathbb N} \in A$, so suppose not, then $\{y_n\}_{n \in \mathbb N}$ is a convergent sequence, so $\{y_n\}_{n \in \mathbb N}-\{a_n\}_{n \in \mathbb N}=\{b_n\}_{n \in \mathbb N}$ is a convergent sequence, which is clearly absurd. This proves that $B(\{a_n\}_{n \in \mathbb N},\epsilon) \cap A \neq \emptyset$

Now it remains to prove $A$ is open, but this is equivalent to prove that $A^c$ is closed: Let $x=\{x_n\}_{n \in \mathbb N}$ be a limit point of $A^c$, I want to show that $x \in A^c$, i.e., that $x$ is a convergent sequence. By hypothesis, there is $\{y^n\}_{n \in \mathbb N}$ (I'm using superscripts here because each $y^n$ is a sequence itself). It's sufficient to show that $x$ is a Cauchy sequence. Here I am not so sure if what I've done is correct, I got confused with so many sequences and inequalities:

Let $\epsilon>0$

By hypothesis, $y^n \to x$, so there is $N : \space \forall \space n \geq N$, $sup_{k \in \mathbb N}|y_{k}^n-x_k|=d_{\infty}(y^n,x)<\dfrac{\epsilon}{3}$.

$|x_k-x_j|\leq |x_k-y_{k}^N|+|y_{k}^N-y_{j}^N|+|y_{j}^N-x_j|<\dfrac{\epsilon}{3}+\dfrac{\epsilon}{3}+|y_{k}^N-y_{j}^N|$, Now, the sequence $\{y_{n}^N\}_{n \in \mathbb N} \in A^c$, which means it is a convergent sequence $\implies$ it is a Cauchy sequence. So, there is $n_0 \in \mathbb N$: if $k,j\geq n_0 \implies |y_{k}^N-y_{j}^N|<\dfrac{\epsilon}{3}$

Then, for all $k,j \geq n_0$, we have that

$|x_k-x_j|\leq |x_k-y_{k}^N|+|y_{k}^N-y_{j}^N|+|y_{j}^N-x_j|<\dfrac{\epsilon}{3}+\dfrac{\epsilon}{3}+\dfrac{\epsilon}{3}=\epsilon$.

We've proved $x$ is a Cauchy sequence $\implies$ it is convergent $\implies x \in A^c$. Since $A^c$ is closed, then $A$ is open.

For part $b)$ I would appreciate some help. I think that $A$ is not separable, I've tried to choose $S \subset A$ such as $S$ the space of sequences of ones and zeros that are not eventually $1$ or eventually $0$, but I got stuck trying to show by absurd that $S$ is not separable. Any ideas or suggestions for this point?

• You could look at what you've already done in part (a). If $A$ is dense in $\ell^\infty$, and $A$ is separable, what does that say about $\ell^\infty$? Alternatively, you could certainly also try to construct uncountably many divergent 0/1-valued sequences in $\ell^\infty$ that all have mutual distance at least one, as you suggest. This approach will work, if you find a clever choice of sequences. – GCD Dec 8 '13 at 22:03
• Hey, thanks! We've proved $A$ is dense in $l_{\infty}$, then $\overline A=l_{\infty}$. Could it be that if a space $S$ is separable, then $\overline S$ is also separable? How could I prove this? If that statement is true, suppose $A$ is separable, then $\overline A=l_{\infty}$ is separable, which is clearly absurd (I can choose $S$ the subspace of sequences of ones and zeros, both convergent and divergent,and it's easy to prove $S$ is not separable, then I would conclude $l_{\infty}$ is not separable by the statement "every subspace of a separable space is separable"). – user100106 Dec 8 '13 at 22:24
• The first thing I've tried was your alternative suggestion, but I couldn't find a clever choice of sequences. – user100106 Dec 8 '13 at 22:27
• For the first approach: In a metric space, if $A\subset B$ is dense in $B$, and $B\subset C$ is dense in $C$, then $A$ is dense in $C$. Try to prove this! It's not too hard, just hammer at the definition of "dense". This should allow you to complete the argument. – GCD Dec 8 '13 at 23:31
• For the second approach, the one you originally tried, try considering binary expansions of real numbers. This gives you a large supply of 0/1-valued sequences. Which ones diverge? Are there uncountably many distinct diverging ones? – GCD Dec 8 '13 at 23:33

a) Openness is usually easier to prove than closedness. If a sequence $(a_n)$ is not Cauchy, then there exists $\epsilon>0$ such that for every $N$ there are $n,m\ge N$ with $|a_n-a_m|\ge \epsilon$. Now if $\|b-a\|_\infty<\epsilon/3$, then with the above indices $|b_n-b_m|\ge \epsilon/3$ by the reverse triangle inequality. Thus, the set of all non-Cauchy sequences is open.
(Note: for real- or complex- valued sequences "non-Cauchy" is equivalent to "non-convergent". But we can imagine a metric space $\ell_\infty(X)$ of bounded sequences in some general metric space $X$; then the openness still holds for non-Cauchy sequences, but may fail for non-convergent ones.)
For density, it suffices to observe that for every $a\in \ell_\infty$ and every $\epsilon>0$ at least one of two sequences $(a_n)$ and $(a_n+(-1)^n\epsilon)$ fails to converge.
b) was settled in comments: since "being dense in" is a transitive relation, a dense subset of $\ell_\infty$ cannot contain a countable dense subset.