If $X$ is normed space, $V \subset X$ is closed, and $W \subset X$ is finite-dimensional with $V \cap W = \{0\}$, then $\pi(V) \subset X/W$ is closed

Asked Dec 27 '22 at 13:23

Active Dec 27 '22 at 13:39

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Let $X$ be a normed space, let $V \subset X$ be a closed subspace, and let $W \subset X$ be a finite-dimensional subspace with $V \cap W = \{0\}$.

I would like to show that $\pi(V) \subset X/W$ is closed, where $\pi : X \to X/W$ is the quotient map.

I was able to show with a proof by contradiction that there is some $C > 0$ such that $$\forall x \in V, \forall y \in W : \|x \| + \|y\| \leq C \|x - y\|. $$

But I don't see how to use the above norm estimate to show that $\pi(V)$ is closed. The estimate above implies that $$\forall x \in V : \|\pi(x)\|_{X/W} \geq \frac{1}{C}\|x\|, $$ and with the additional fact that $\pi$ is bounded, we have $$\forall x \in V: \frac{1}{C}\|x\| \leq \|\pi(x)\|_{X/W} \leq \|x\|, $$ but I cannot conclude that $\pi(V)$ is closed without knowing that $V$ is complete, which a priori is not.

edited Dec 27 '22 at 13:25

asked Dec 27 '22 at 13:23

user1134586

The seminorm on $X/W$ is a norm if $W$ is closed, so that's where one can invoke finite dimension. – AlvinL Dec 27 '22 at 13:54
@AlvinL Indeed, since $W$ is finite-dimensional, it is closed in $X$, so $X/W$ is a normed space, but this is as much as I can say about it. – user1134586 Dec 27 '22 at 13:55
Might be of help (somehow) – AlvinL Dec 28 '22 at 10:22

If $X$ is normed space, $V \subset X$ is closed, and $W \subset X$ is finite-dimensional with $V \cap W = \{0\}$, then $\pi(V) \subset X/W$ is closed

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