Riesz's Lemma for finite-dimensional spaces Let $Z$ be any linear normed space and $Y$ a closed subspace of $Z$, then by Riesz's Lemma there exist an element $z_\theta \in Z$ such that $||z_\theta||=1$ and  $dist(z_\theta,Y)<\theta$ where $\theta\in (0,1)$, i.e we can get away from the closed subspace arbitrarily close to $1$ and still be on the unit ball.
When however, $Z$ is a finite dimensional linear normed space, one can find an element $z\in Z$ such that $\operatorname{dist}(z,Y)=1$ with $\|z\|=1$. Could someone give me an idea of ​​what happens in finite dimensional case?
 A: This is a comment to David Mitra's answer. There is a beautiful result by Krein, Milman and Krasnoselski based on the Borsuk–Ulam theorem which strengthens the Riesz lemma in finite dimensions.

Theorem. Let $M$ and $N$ be finite-dimensional subspaces of a normed space. If $\dim M > \dim N$, then there is $x_0\in M$ such that ${\rm dist}(x_0, N)=\|x_0\|$.

I am happy to provide a proof on request. I am very curious whether there exists a proof of this statement which would avoid using the Borsuk–Ulam theorem.
Reference:


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*M. Kreĭn, M. Krasnoselski, D. Milman, On defect numbers of operators on Banach spaces and related geometric problems, Trudy Inst. Mat. Akad. Nauk Ukrain. SSR, 11 (1948), 97–112.

A: Since the following result doesn't seem to be well-known, I'll post its statement and an outline of its proof, both  taken from the Lemma mentioned in the link in my comment above:
Theorem: 
Let $Y$ be a finite dimensional subspace of the normed vector space $X$. Then there is an element $x\in X$ of norm one with $\text{dist}\,(x,Y)=1$.

Outline of proof: 
Let $z\in X\setminus Y$. Then, as $Y$ is closed,  $t=\text{dist}\,(z, Y)\ne0$. One can show, using the fact that $t^{-1}Y=Y$,  that
$$\text{dist}\,(t^{-1}z,Y)=1.$$ Let $(y_n)$ be a sequence in $Y$ with 
$$\Vert y_n-t^{-1}z\Vert\rightarrow 1\tag1$$ Then $(y_n)$ is bounded and as $Y$ is finite dimensional, there is  a subsequence $(y_{n_k})$ of $(y_n)$ that converges to some $y\in Y$.  
Let $x=y-t^{-1}z$.
One can use $(1)$ to show that $\Vert x\Vert =1$.
Using the fact that $y-Y=Y$, one can show that $\text{dist}\,(x, Y)=1$.
Thus, $x$ is as advertised.
