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?
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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.
answered Apr 29, 2020 at 4:19
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$\begingroup$ Why there is some $z\in N$ with $d(y,z)<r$? This is the point I cannot prove. $\endgroup$ Commented Apr 29, 2020 at 4:21
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$\begingroup$ @flyingwith Given any bounded non-empty set $A\subset \mathbb{R}$, if $\inf A <r$, then $r$ is not a lower bound of $A$ (because $\inf A$ is the largest lower bound), which means that there exists some $a\in A$ with $a<r$. $\endgroup$ Commented Apr 29, 2020 at 4:23
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$\begingroup$ Could you also say the infimum is attained in $N$ because it is closed? $\endgroup$– foam78Commented Apr 29, 2020 at 4:36
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$\begingroup$ @Connor_Tracy No, it false for general Banach spaces (it holds in Hilbert spaces though, by taking orthogonal projections). Here's a counterexample: Let $X:=\{f\in C([0,1]): f(0)=0$. Let $\phi(f):=\int_0^1 f$ (wrt to Lebesgue measure), $N=\ker \phi$ (you can check that $\phi$ is continuous). Assuming you have a function $f\in X$ with $d(f, N)=1=\|f\|$, you can argue that $|\phi(f)|\|g\|\ge |\phi(g)|$ for every $g\in X$. If you consider $f_n(t)=t^\frac{1}{n}$, then $f_n\in X\setminus N$, and you can show using the previous fact that $|\phi(f)|\ge 1$. But $f\in M$, so $f(0)=0$. $\endgroup$ Commented Apr 29, 2020 at 4:45
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$\begingroup$ @Connor_Tracy Therefore, using that $\|f\|_\infty =1$, $\phi(f) \le \int _0^1 |f| <1$, contradiction. $\endgroup$ Commented Apr 29, 2020 at 4:51