# Distance between a normed vectorspace and a finite dimensional subspace

I need to prove the following statement:

"$X$ is a normed vectorspace and $Y$ is a finite dimensional subspace. For all $x \in X$ the distance is defined as $d(x,Y)=\inf\{ \|x-y\|:\ y \in Y\}$"

So I need to prove that there is an specific y in Y such that $d(x,Y) = \|x-y\|$.

I used the Corollary that every finite dimensional subspace of a normed vectorspace is also closed, and then I tried to apply the Riesz Lemma. But I am not sure.

Can anyone give me a hint for this exercise?

• In a finite-dimensional normed vector space (or subspace) every closed bounded subset is compact. Any $n$-dimensional real normed vector space has a linear homeomorphism to $\mathbb R^n.$ – DanielWainfleet Oct 9 '16 at 16:51

## 1 Answer

Use the definition of infimum to find a sequence $\left(y_n\right)_{n\geqslant 1}$ of elements of $Y$ such that $\lVert x-y_n\rVert\leqslant d(x,Y)+1/n$ for each $n\geqslant 1$. The sequence $\left(y_n\right)_{n\geqslant 1}$ is bounded (why?) and lies in a normed vector space of finite dimension: you can extract a converging subsequence.

• What exactly is that showing me? Sorry I don't get the point of doing that,.. – Yuhe Oct 9 '16 at 9:40
• The wanted $y$ will be the limit of the converging subsequence. – Davide Giraudo Oct 9 '16 at 9:41
• Ah I see, thanks :) the sequence is bounded because it is in a finite space I guess? or is there a better reason? – Yuhe Oct 9 '16 at 9:52
• No, juse use $||y_n||\leqslant ||y_n-x|| +d(x,Y) \leqslant 1/n+||x|| \leqslant 1+||x ||$. – Davide Giraudo Oct 9 '16 at 9:56
• How do you get to this inequality? – Yuhe Oct 9 '16 at 10:38