What are the interior points of this set $A=\{(x_n)\in \ell_2 \mid |x_n|<\frac1n, n=1,2,3\ldots\}$?

Question

I know for one that $0=(0,0,\dots) \in A$ is not an interior point.

because if $\exists \epsilon>0$ such that $B_{\epsilon}(0) \subseteq A$ then I can get $n_0>\frac2\epsilon$ such that if $x$ is the sequence that has $\frac\epsilon2 $ at its $n_0$ position and $0$ everywhere else then this $x$ lies in $B_{\epsilon}(0)$ because $\|x\|_2 =\frac\epsilon2$, but $x$ is not in $A $ since $|x_{n_{0}}|= \frac \epsilon2$ $>\frac1{n_0}$

How do I find other interior points?

Hint: for $u \in A$, $\epsilon>0$, consider $u_n +\epsilon n^{-0.75}$. — Aphelli, Sep 19 '19 at 20:48
Can I somehow show that it is subspace of $\ell 2$ and then it will have no interior point? — chesslad, Sep 19 '19 at 20:49
$A$ is a subspace of the sub-vector space $A’={x \in \ell^2,, (nx_n) \in \ell^{\infty}}$. — Aphelli, Sep 19 '19 at 20:55

score 1 · Accepted Answer · answered Sep 19 '19 at 21:14

The set $A$ has empty interior. Prove by contradiction. Suppose that $A$ has an interior point $a$, then there exists $\varepsilon>0$ such that $B(a;\varepsilon)\subseteq A$. Define a sequence $(b_{n})$ in $l^{2}$ by $$ b_{n}(k)=\begin{cases} 0, & \mbox{if }k\neq n\\ \frac{\varepsilon}{2\sqrt{n}}, & \mbox{if }k=n\mbox{ and }a(n)\geq0,\\ -\frac{\varepsilon}{2\sqrt{n}}, & \mbox{if }k=n\mbox{ and }a(n)<0. \end{cases} $$ Clearly $||b_{n}||=\frac{\varepsilon}{2\sqrt{n}}<\varepsilon$, so $a+b_{n}\in B(a;\varepsilon)\subseteq A$. In particular, $|(a+b_{n})(n)|<\frac{1}{n}$. Note that $a(n)$ and $b_{n}(n)$ are of the same sign, so \begin{eqnarray*} |(a+b_{n})(n)| & \geq & |b_{n}(n)|\\ & = & \frac{\varepsilon}{2\sqrt{n}}. \end{eqnarray*} Hence, we obtain $\frac{\varepsilon}{2\sqrt{n}}<\frac{1}{n}$ for all $n\in\mathbb{N}$, which is a contradiction.

What are the interior points of this set $A=\{(x_n)\in \ell_2 \mid |x_n|<\frac1n, n=1,2,3\ldots\}$?

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