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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?

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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.

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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 $|x|_s=\epsilon$.

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  • $\begingroup$ 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 $x\not \in B_ {|.| _1} (0,1)$ $\endgroup$ – eraldcoil Dec 19 '18 at 17:24
  • $\begingroup$ I'm sorry, but I do not understand what you are trying to say. $\endgroup$ – SmileyCraft Dec 19 '18 at 17:25
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Because every continuous open bijection is an homeomorphism.

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