Quotient of a Banach space $X$ gets quotient topology under standard norm induced from $X$.

Question

Let $X$ be a Banach space and $Y$ be a closed subspace of $X$. Then there is a canonical norm on $X/Y$. I want to show that this norm induces quotient topology on $X/Y$. Any hint/solution? I was trying to show that $\pi :X\to X/Y$ is open, so we need to show that $\exists$ $\epsilon >0$ s.t. $B_{X/Y}(0,\epsilon) \subset \pi(B_{X}(0,1)).$ Can someone please help?

Answer 1

First, if $x\in X$, then $$ \|\pi(x)\|_{X/Y}=\|x+Y\|_{X/Y}=\inf_{y\in Y}\|x+y\|_X\le \|x\|_X, $$ and hence $\pi$ is bounded.

Next, as $$ \pi\big(B_X(0,1)\big)=\left\{x+Y:x\in B_X(0,1)\right\}, $$ let $\hat z\in B_{X/Y}(0,1)$. Then $\hat z=z+Y$, with $$ \|\hat z\|=\|\pi(z)\|_{X/Y}=\|z+Y\|_{X/Y}=\inf_{y\in Y}\|z+y\|_X<1, $$ and hence there exists a $y\in Y$, such that $\|z+y\|_X<1$, i.e., $z+y\in B_X(0,1)$, and hence $$ \hat z=\pi(y+z)\in \pi\big(B_X(0,1)\big). $$ Thus $$ B_{X/Y}(0,1)\subset \pi\big(B_X(0,1)\big), $$ and hence $\pi$ is open.