Is $c_0$ meagre in $l^\infty$?

Question

I have been reading up on the meagre sets sans an instructor after a 10 year break from this sort of mathematics and I would like to test my understanding. I want to know if the following is true (and if so, if anyone has a nice proof):

Is $c_0$ meagre in $l^\infty$ ? (where $c_0$ is the subset of $l^\infty$ where for $x_n$=the nth element of x, we have$lim_{n\to\infty} x_n=0$)

My approach is as follows (please correct me if I am wrong):

A space is meagre if it can be written as a countable union of nowhere dense sets. And a set is nowhere dense if it is closed and has no interior points.

So one needs to show that

1. $c_0$ is a subspace of $l^\infty$.

2. $c_0$ is closed in $l^\infty$.

3. The interior of $c_0$ is empty.

Then one could say that $c_0$ is nowhere dense and of course (?) it could be written as a countable union of nowhere dense sets (just itself?), so it is meagre.

For 1:

$c_0$ is non-empty as x=(1,0,0,0,...) is an element of $c_0$.

Let $x$ and $y$ be elements of $c_0$. This means $||x||_\infty=sup_n(|x_n|)<\infty$ and same for $y$. Then

$||x+y||_\infty=sup_n(|x_n+y_n|)=sup_n(|x_n|)+sup_n(|y_n|)<\infty$

and

$lim_{n\to\infty} (x_n+y_n)=lim_{n\to\infty} x_n+lim_{n\to\infty} y_n=0$

and

$||ax||_\infty=sup_n(|ax_n|)=|a|sup_n(|x_n|)<\infty$

and

$lim_{n\to\infty} (ax_n)=a lim_{n\to\infty} x_n=0$

so $x+y$ and $ax$ are elements of $c_0$. Thus, $c_0$ is subspace of $l^\infty$.

For 2:

Let $x^m$ be an element of $c_0$ and $x$ be an element of $l^\infty$.

$|x_n| \leq |x_n-(x^m)_n|+|(x^m)_n|$

$|x_n-(x^m)_n| \to 0$ (as $x \in l^\infty$) and $|(x^m)_n| \to 0$ (as $x^m \in c_0$)

so $|x_n| \to 0$ and then $x \in c_0$ and then $\overline c_0=c_0$ so $c_0$ is closed.

For 3:

HELP!!

Is my approach a workable approach? And/or is there a better one?

Answer 1

An idea how to tackle step 3: If a $x\in\mathring c_0$, then there is a ball of sufficiently small radius around $x$ that is still in $c_0$. So it is sufficient to show that for every $x\in c_0$ and every $\varepsilon>0$, there exists an element $a\in l^\infty$ such that $\|a-x\|_{l^\infty}<\varepsilon$ but $a\in l^\infty\setminus c_0$.

Proof: Let $x=(x_1,x_2,\ldots)\in c_0$ and $\varepsilon>0$ be fixed. There exists an index $N$ such that $|x_n|<\frac{\varepsilon}{2}$ for all $n\geqslant N$. Now set $$ a:=\left(x_1,\ldots,x_N,\frac{\varepsilon}{2},\frac{\varepsilon}{2},\ldots\right). $$ Clearly, $a\in l^\infty\setminus c_0$ and $\|a-x\|_{l^\infty}<\varepsilon$.

EDIT: daws answer is somewhat cleaner than mine (but the idea is the same).

Answer 2

For 3: The idea is to perturb an element $x\in c_0$ in such a way that the perturbed element is not in $c_0$ anymore, i.e. its limit does not exist or is non-zero.

Take $x \in c_0$. Take the constant sequence $e=(1,1,\dots)\in l^\infty$. Now let $\delta>0$. Then $$ x + \delta e \not\in c_0, $$ since $\lim_{n\to\infty}(x_n + \delta e_n)=\delta\ne 0$. Moreover, $$ \|x-(x+\delta e) \|_\infty= \|\delta e \|_\infty=\delta, $$ which implies that in every open ball around $x$ there is another element $x+\delta e$ such that $x+\delta e\not\in c_0$. Hence $c_0$ has no interior point.