A non trivial counterexample in three-space property

Question

We know that completeness is a three-space property: Let $M$ be a closed subspace of a normed space $X$. Then, $X$ is complete if and only if $M$ and $X/M$ are complete.

I am looking for a non trivial counterexample: $X$ incomplete normed space but $X/M$ ($\neq \{M\}$) complete space with $M$ closed subspace of $X$.

score 4 · Answer 1 · answered Sep 13 '15 at 06:00

4

Let $M$ be any incomplete normed space and $Y$ be any complete normed space. Let $X=M\oplus Y$ (with, say, the norm $\|(m,y)\|=\|m\|+\|y\|$). Then $M$ is a closed subspace of $X$ and the quotient $X/M\cong Y$ is complete, but $X$ is not complete because $M$ is not.

answered Sep 13 '15 at 06:00

Eric Wofsey

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Dear Sir,Why M is closed subspace of X? – Curious student Aug 11 '19 at 15:27
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If $(m,y)\not\in M$ then $y\neq 0$ so its distance from any element of $M$ is at least $|y|>0$. – Eric Wofsey Aug 11 '19 at 15:58

A non trivial counterexample in three-space property

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