Open sets in Lebesgue sequence spaces

Question

Is $\{x\in \ell^1:\|x\|_1<1\}$ open in $(\ell^1,\|x\|_2)$? In case of $\mathbb R^n$ it is definitely "yes" as any two norms are equivalent there. In this case we know that $\|x\|_2\leq \|x\|_1$ for all $x\in \ell^1$. But what next? I am confused.

Answer 1

(0). I am using $l^1$ for the set of absolutely summable sequences and $l_1,l_2$ for the norms and their associated topologies.

(1). If $d_1$ and $d_2$ are metrics on the same set $X$ and there exists $k>0$ such that $d_2(p,q)\leq k\cdot d_1(p,q)$ for all $p,q\in X$ then the topology generated by $d_1$ is stronger than (or equal to ) the topology generated by $d_1.$

With $k=1$ and $d_i(p,q)=\|p-q\|_i$ for $p,q\in l_1$ and $i\in \{1,2\},$ therefore the $l_1$-topology on the set $l^1$ is stronger (or equal to) the topology generated by the $l_2$ norm.

(2). For brevity let $A=\{x\in l^1:\|x\|_1<1\}.$ Observe that any $S\subset l^1$ is $l_2$-open iff $y+rS =\{y+rx:x\in S\}$ is $l_2$-open for all $r>0$ and all $y\in l^1.$

So if $A$ is $l_2$-open then every $l_1$-open ball is $l_2$-open , because the $l_1$-open ball $B_1(y,r)$ (with $y\in l^1$ and $r>0$) is equal to $y+rA.$ This would imply that the $l_2$-topology on the set $l^1$ is stronger than or equal to the $l_1$ topology.

(3). By (1) and (2) if $A$ is $l_2$-open then the $l_1$ and $l_2$ norms generate the same topology on the set $l^1$. But consider $x_i=(x_{i,n})_{n\in N}$ where $x_{i,n}=(i\ln (n+1))^{-1}$ for $i\leq n,$ and $x_{i,n}=0$ for $i>n.$ Let $y=(y_n)_{n\in N}$ where $y_n=0$ for every $n.$

Then $y\in l^1$ and every $l_2$-nbhd of $y$ contains $x_i$ for infinitely many $i\in N$ because $\lim_{i\to \infty }\|x_i-y\|_2=0.$ But $\lim_{i\to \infty}\|x_i\|_1=1$ and $\|y\|_1=0$ so the $l_1$-open ball $\{z\in l^1:\|z\|<1/2\}$, which is an $l_1$-nbhd of $y$, contains $x_i$ for only finitely many $i\in N.$ So the topologies on $l^1$ generated by the two norms are not identical. So $A$ is not $l_2$-open.