The openness of positive sequences in $\ell^{2}$

Question

Let $S=\{\{x_n\}\in \ell^2|\forall n\;x_n>0\}$. I have to evaluate whether or not S is open. I believe it's not the case. However, I think my proof includes fallacious reasoning and would appreciate some feedback.

Consider the sequence $x_n=\frac{1}{n}$, which is in $\ell^2$ and, $x_n >0$ $\forall n$. Now, in $\mathbb{R}$, we have that $\frac{1}{n}\to 0$. Thus, $\forall \varepsilon>0, \exists N \in \mathbb{N}$ s.t. $m>N \implies \left\lvert \frac{1}{m}\right\rvert <\varepsilon$. Let's now consider the sequence

$$y_n=\begin{cases} \frac{1}{n}-\frac{1}{m}, & \text{if $n=m+1$} \\ \frac{1}{n}, & {\forall n \ne m+1 } \end{cases}$$

Again, $y_n \in \ell^2$.

Let $r>0$ be arbitrary. We have that $y_n \in B_{r}(x_n)$. Indeed, $$d(x_n, y_n)=\sqrt{\sum_{i=0}^\infty \left(x_n - y_n\right)^2 }=\frac{1}{m}$$

and, for all radii r, we can chose the appropriate sequence $y_n$ such that $\frac{1}{m}<r$. However, when $n=m+1$, $y_n \ngtr 0$. Thus, $y_n \notin S$, and $B_{r}(x_n) \nsubseteq S$. Since r was arbitrary, we conclude that S is not open.

I have a feeling that, since $y_n$ depends on the radius, my proof is incorrect. Any comments? Corrections? Am I totally offtrack?

Answer 1

Looks fine to me. A rephrasing of your proof: $1/n$ for large $n$ is tiny, so for any $\epsilon>0$, you can poke it at the $m$th coordinate for $m\gg 1$ by $\epsilon$ to get $$y = x - \epsilon \mathbf e^m,$$ where $(\mathbf e^m)_n = \delta_{mn}$ is the 'standard basis vector', so that the $m$th coordinate of $y$ is negative. Notice $m=m(\epsilon)$, so there is no $\epsilon$ ball around $x$ for any $\epsilon>0$.

Answer 2

Your proof is correct, you have shown that for $x = \left(\frac1n\right)_n \in S$ there does not exist $r > 0$ such that the open ball $B(x, r)$ is contained in $S$. This means that $S$ is not open.

Another approach would be to show that $S^c = \{(x_n)_n \in \ell^2 : x_n \le 0 \text{ for some } n \in \mathbb{N }\}$ is not closed in $\ell^2$.

Indeed, again take $x = \left(\frac1n\right)_n \in S$ and define a sequence $\left(y^{(n)}\right)_n$ of vectors in $S^c$ defined as

$$y^{(n)}_k = \begin{cases} \frac1k, & \text{if $k\ne n$} \\ 0, & \text{if $k = n$} \end{cases}$$

We have that $$\left\|x - y^{(n)}\right\|_2 = \left|x_n - y^{(n)}_n\right| = \frac1n \xrightarrow{n\to\infty} 0$$

so $x = \lim_{n\to\infty} y^{(n)}$. We conclude that $S^c$ cannot be closed because we have found a convergent sequence in $S^c$ whose limit is not in $S^c$.