Set of convergent sequences is not dense in $(\ell^\infty, d_\infty)$

Question

Let $c_c$ be the set of all sequences which converge with respect to the usual metric on $\mathbb{R}$. Since every convergent sequence is bounded, then we know that $c_c\subseteq \ell^\infty$, but would we show that $c_c$ is not dense in $(\ell^\infty,d_\infty)$? By definition, a subset $D$ of $X$ is dense in $X$ if $\overline{D}=X$, but what is the closure of $c_c$?

Answer 1

Since $\ell^\infty$ is a normed vector space, let us use the norm instead of the metric for convenience.

$c_c$ is not dense in $\ell^\infty$; consider $ y =(1,0,1,0,\dots) \in \ell^\infty$. If $$ \| y - x \|_\infty<1/2$$ this implies $x(2n+1) > 1/2$ but also $x(2n)<1/2$, for large enough $n$. Consequently, there is no element of $c_c$ in the $\ell^\infty$ ball around $y$ of radius $1/2$.

$c_c$ is a closed space; suppose $x_n \in c_c$ and $x_n \to x \in \ell^\infty$. Then \begin{align} |x(k) - x(j)| &\le |x(k) - x_n(k)| + |x_n(k) - x_n(j)| + |x_n(j) - x(j)| \\&\le 2\|x_n-x\|_\infty + |x_n(k) - x_n(j)| \end{align}

So for any $\epsilon>0$, we can choose a large $n$ so that$ \|x_n-x\|_\infty < \epsilon $ and then choose $M$ so that if $k,j>M$ then $|x_n(k) - x_n(j)|<\epsilon$. Thus, $x$ is Cauchy and thus convergent.

Answer 2

The closure of $c_c$ is again $c_c$. In other words, $c_c$ is a closed set. It is easy to show that $\ell^\infty\setminus c_c$ is open using that fact that a sequence doesn't belong to $c_c$ if and only if it is not a Cauchy sequence.