# Open mappings

Gold Member

## Homework Statement

Let $(X,||\cdot||)$ be a normed vector space and suppose that $Y$ is a closed vector subspace of $X$. Show that the map $||x||_1=\inf_{y \in Y}||x-y||$ defines a pseudonorm on $X$. Let $(X/Y,||\cdot||_1)$ denote the normed vector space induced by $||\cdot||_1$ and prove that the canonical projection $\phi:X \rightarrow X/Y$ is an open mapping.

## Homework Equations

All vector spaces are over $\mathbb{R}$.

## The Attempt at a Solution

So far I have been able to show that $||\cdot||_1$ is a pseudonorm, but I am having difficulty showing the canonical projection is an open mapping. Obviously we need to take $U$ open in $X$ and then take $[x] \in \phi(U)$. From here we need to construct an open neighborhood around $[x]$ which is contained in $\phi(U)$ but I am having difficulty doing this. Any help?

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AKG
Homework Helper
What does the fact that $x\in U$ with $U$ open tell you? Based on what that gets you, you should be able to think of a candidate for the desired open neighbourhood containing $[x]$ and contained in $\phi(U)$.

Gold Member
Well I know that there exists $0 < \varepsilon$ such that $B(x,\varepsilon) \subseteq U$. But the problem I then have is that $||\cdot||_1 \leq ||\cdot||$ so I am having trouble finding a criterion with $||x-y||_1 < \delta$ implies $||x-y|| < \varepsilon$.

We wish to prove that $B_1(\varphi(x),\varepsilon)\subseteq \varphi(B(x,\varepsilon))$. We can take x=0 if we want.

So take $z\in B_1(0,\varepsilon)$, write out what that means.