Linear continuous bijection but not open.

Question

I have the next question. Let $l^1$ be the set of sequences $(a_1,a_2,\ldots, )$ such that $\sum |a_k|<\infty$. If we consider norm $|.|_1$ and the supremum norm $|.|_{s}$, then $(l^1,|.|_1)$ is complete , while $(l^1,|.|_s)$ is not complete.

Let $id:(l^1,|.|_1)\to (l^1,|.|_s),\ x\to x$ is a continuous bijection but is not open.

Why is not open?

Answer 1

Because every continuous open bijection is an homeomorphism.

Answer 2

If a bijection between topological spaces is open, then its inverse is continuous. In your case this would imply that $\vert x\vert_1 \le C \vert x\vert_s$ for a uniform constant $C$. Consider the sequence $x^n=(1,\dots,1,0,\dots)\in l^1$ (ones in the first $n$ entries and zeros afterwards) to see that this is wrong.

Answer 3

Consider the set $B:=\{x\in\ell^1:|x|_1<1\}$, which is open with respect to $(\ell^1,|\cdot|)$. We find $\mbox{id}B=B$, however $B$ is not open with respect to $(\ell^1,|\cdot|_s)$, since $0$ is an element, but not an interior point.

Proof: Let $\epsilon>0$ and assume without loss of generality that $\epsilon<1$. Then $x\in\ell^1$ defined by $x_n=\epsilon$ if $n\epsilon<2$ and $x_n=0$ otherwise. Then $|x|_1>1$, however $|x|_s=\epsilon$.