# Rudin Functional Analysis, Lemma 4.22

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:

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, $$\inf_{y \in M}\left\|\frac{x}{\lambda}-y\right\|= \inf_{y \in M}\left\|\frac{x}{\lambda}- \frac{y}{\lambda}\right\|= \frac{1}{\lambda} \inf_{y \in M}\|x-y\| = 1.$$
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 $$1$$.