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3 Corrected two typos.

This is just to make Nate Eldridge's answer selfcontained.

For any normed vector space $V$ and any $r > 0$, write $V_r$ for the open ball of radius $r$ and center $0$ in $V$.

Let $X$ be a normed vector space, $Y$ a closed space, $Z$ the quotient, $\pi$ the canonical projection, $\tau$ the quotient topology, $\nu$ the topology on $Z$ induced by the quotient norm $$|\pi(x)|:=\inf_{y\in Y}|x+y|.$$

(It's easy to see that this is a norm.)

We claim $\tau=\nu$.

Both topologies are translation invariant.

The set ${\pi(X_r)\ \{\pi(X_r)\ |\ r > 0}$ 0\}$is a basis for the$\tau$-neighborhoods of$0$in$Z$. The set${(Z_r)\ \{(Z_r)\ |\ r > 0}$0\}$ is a basis for the $\nu$-neighborhoods of $0$ in $Z$.

As $\pi(X_r)=Z_r$, we're done.

This is just to make Nate Eldridge's answer selfcontained.

For any normed vector space $V$ and any $r > 0$, write $V_r$ for the open ball of radius $r$ and center $0$ in $V$.

Let $X$ be a normed vector space, $Y$ a closed space, $Z$ the quotient, $\pi$ the canonical projection, $\tau$ the quotient topology, $\nu$ the topology on $Z$ induced by the quotient norm $$|\pi(x)|:=\inf_{y\in Y}|x+y|.$$

(It's easy to see that this is a norm.)

We claim $\tau=\nu$.

Both topologies are translation invariant.

The set ${\pi(X_r)\ |\ r > 0}$ is a basis for the $\tau$-neighborhoods of $0$ in $Z$.

The set ${(Z_r)\ |\ r > 0}$ is a basis for the $\nu$-neighborhoods of $0$ in $Z$.

As $\pi(X_r)=Z_r$, we're done.

1

This is just to make Nate Eldridge's answer selfcontained.

For any normed vector space $V$ and any $r > 0$, write $V_r$ for the open ball of radius $r$ and center $0$ in $V$.

Let $X$ be a normed vector space, $Y$ a closed space, $Z$ the quotient, $\pi$ the canonical projection, $\tau$ the quotient topology, $\nu$ the topology on $Z$ induced by the quotient norm $$|\pi(x)|:=\inf_{y\in Y}|x+y|.$$

We claim $\tau=\nu$.

Both topologies are translation invariant.

The set ${\pi(X_r)\ |\ r > 0}$ is a basis for the $\tau$-neighborhoods of $0$ in $Z$.

The set ${(Z_r)\ |\ r > 0}$ is a basis for the $\nu$-neighborhoods of $0$ in $Z$.

As $\pi(X_r)=Z_r$, we're done.