# Linear continuous bijection but not open.

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?

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.
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 $$|x|_s=\epsilon$$.
• Excuse. $B$ is open with respect to $(l^1|.|_1)$ and $B$ is not open with respect to $(l^1,|.|_s)$? I ask it since in the proof, $x\in B_ {|. | _ {s}} (0,1)$ and $x\not \in B_ {|.| _1} (0,1)$ – eraldcoil Dec 19 '18 at 17:24