Proof of Theorem 1.41(d) on Rudin's Functional Analysis. Inducing an invariant metric on the quotient space $X/N$.

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I have a question from the proof of Theorem 1.41 (d) in Rudin's Functional Analysis.

Namely, let $N$ be a closed subspace of a topological vector space $X$. Let $\tau$ be the topology of $X$ and define $\tau_N$ as the quotient topology on $X /N$.

In this case, 1.41 (d) shows that if $\tau$ is induced by a complete invariant metric $d$, then $\tau_N$ is also induced by some complete invariant metric $\rho$.

Proof: Suppose that $d$ is an invariant metric on $X$, compatible with $\tau$. Define $\rho$ by $$\rho(\pi(x),\pi(y))= \inf\{d(x-y,z):z \in N\}.$$ This may be interpreted as the distance from $x-y$ to $N$. We omit the verifications that are now needed to show that $\rho$ is well defined and that it is an invariant metric on $X/N$. Since $$\pi(\{x:d(x,0)<r\})=\{u:\rho(u,0)<r\},$$ it follows from (b) that $\rho$ is compatible with $\tau_N$.

In this proof, how do we show that $\rho$ is well-defined invariant, i.e., if $x_1 - x_2 \in N$ and $y_1-y_2 \in N$, then $d(x_1-y_1,N)=d(x_2-y_2,N)$?

Also, how do we show that $\pi(\{x:d(x,0)<r\})=\{u:\rho(u,0)<r\}$?

I would greatly appreciate any help with these questions.

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I think I have proved this.

Well-defined : Suppose we have $x_1 - x_2 \in N$ and $y_1 - y_2 \in N$.

Then for any $z \in N$, $d(x_1 - y_1, z)=d(x_1 - y_1 + x_2 - y_2, z + x_2 - y_2) = d(x_2 - y_2, z+ x_2 - y_2 - x_1 + y_1)$.

Now, on the right, $x_2-y_2 - x_1 + y_1 \in N$, and $z$ varies over the linear subspace $N$, so by taking the infimum on both sides over $z \in N$, we get $d(x_1-y_1,N)= d(x_2-y_2,N)$.

$\pi\{x:d(x,0)<r\} = \{u: \rho (u,0)<r\}$:

Suppose $d(x,0)<r$ for some $x \in X$. Then $\rho(\pi(x),\pi(0)) = d(x,N)$. This is $0$ if $x \in N$ and if $x \notin N$, we have $d(x,N) \le d(x,0) < r$. So we have $\rho(\pi(x),0)<r$.

Next, suppose $\rho(u,0)<r$. Then $u = x + N = \pi(x)$ for some $x \in X$. This means that for some $n \in N$, we have $d(x,n)<r$. By invariance, $d(x-n,0)<r$. But $\pi(x-n)=\pi(x) - \pi(n) = \pi(x) = u$. So we get the left inclusion as well.