Rudin's Functional Analysis Theorem 1.41

Question

While I was reading the proof of Theorem 1.41 in Rudin's Functional Analysis, I was stuck in the equation \begin{equation*} \pi(\{x:d(x,0)<r\}) = \{u:\rho(u,0)<r\} \end{equation*} where $N$ is a closed subspace of a topological vector space $X$, $\pi$ is the quotient map of $X$ onto $X/N$, $d$ is a complete translation-invariant metric on $X$, and $\rho(\pi(x),\pi(y)) = \inf\{d(x-y,z):z\in N\}$. Specifically, I cannot prove $\{u:\rho(u,0)<r\}\subset \pi(\{x:d(x,0)<r\})$. I guess that it seems necessary to show that for every $x\in X$ there exists $y\in N$ satisfying $d(x,y) = \inf\{d(x,z):z\in N\}$, but not sure. Would you give me any hint?

Answer 1

Fix $u$ with $\rho(u,0)<r$, say $u=\pi(y)$. Since $$ \inf \{d(y,z):z\in N\}<r $$ there is some $z\in N$ with $d(y,z)<r$. Then $d(y-z,0)<r$ and $$\pi (y-z)=\pi(y)-\pi(z)=\pi(y)=u,$$ so taking $x=y-z$ does the trick.