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I am having difficulty understanding Rudin's proof on Lemma 4.22 of his Functional Analysis book.

The assumption is that $M$ is a subspace of a normed space $X$ and $M$ is not dense in X. Rudin then claims that there exists $x_1 \in X$ whose distance from $M$ is 1, that is, $\inf \{||x_1 - y||: y \in M\} = 1$.

It is not so obvious to me why such an $x_1$ exists. May someone explain the logic to me? Does it have anything to do with the assumption "$M$ is not dense in X"?

Thanks in advance to everyone who's trying to help out.

This is the screenshot of the whole Lemma and proof given by Rudin: enter image description here

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Since $M$ is not dense in $X$, there is a vector $x \in X$ with $\lambda:= \inf_{y \in M}\|x-y\| > 0$. Indeed, assume to the contrary that for every $x \in X$, we have $\inf_{y \in M}\|x-y\| =0$, then we can find a sequence $\{y_n\}\subseteq M$ such that $\|x-y_n\| \to 0$, which means that $M$ is dense in $X$. A contradiction.

Then, since $M$ is a subspace, we have $\lambda^{-1}M = M$ and thus, $$\inf_{y \in M}\left\|\frac{x}{\lambda}-y\right\|= \inf_{y \in M}\left\|\frac{x}{\lambda}- \frac{y}{\lambda}\right\|= \frac{1}{\lambda} \inf_{y \in M}\|x-y\| = 1.$$

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Sense $M$ is not dense, there is some open ball $B_r(x_0)$ in $X$ such that $B_r(x_0)\cap M=\emptyset$. So, for each $m\in M$, $\|m-x\|\geqslant r$. Let $d$ be the distance from $x$ to $M$. Then $d\geqslant r>0$. But then, for each $m\in M$,$$\left\|m-\frac1dx\right\|=\frac1d\|dm-x\|\geqslant1.$$So, the distance from $\frac1dx$ to $M$ is at least $1$. Now, take a sequence $(m_n)_{n\in\Bbb N}$ of elements of $M$ such that $\lim_{n\to\infty}\|m_n-x\|=d$. Then $\lim_{n\to\infty}\left\|\frac1dm_n-\frac1dx\right\|=1$. It is proved then that the distance from $\frac1dx$ to $M$ is equal to $1$.

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