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The Riesz' Lemma is as follows:

Let $Y$ and $Z$ be subspaces of a normed space $X$ of any dimension (finite or infinite) such that $Y$ is closed (in $X$) and is also a proper subset of $Z$. Then for every real number $\theta$ in the open interval $(0,1)$, there is a point $z$ in $Z$ such that $$||z|| = 1$$ and $$||z-y|| \geq \theta$$ for every $y$ in $Y$.

Now we want to prove the following assertion:

If $Y$ is finite-dimensional, then we can even take $\theta$ to be equal to $1$ in the statement of the Riesz' Lemma.

How to prove this?

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What is this space $Z$ for? It's not in the formulation of the Riesz lemma I have and I think it's unncessary. If you want to use this space $Z$, you can get rid of $X$ and say "$Y$ is a closed subspace of $Z$ ... or you drop $Z$ and instead of $z\in Z$ you say that there is some $x\in X$ for which ... holds. –  Elmar Zander Feb 15 '13 at 10:23
    
Well, in a way you're right, but stated this way the result seems a bit more general than otherwise. –  Saaqib Mahmuud Feb 15 '13 at 20:05
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I would apply the following trick in case $\dim Y<\infty$:

Let $z_0\in Z\setminus Y$ arbitrary and imagine the finte dimensional normed space $U$ spanned by $\langle Y,z_0\rangle$. Then there are many ways to continue, for example, the unit ball is compact in $U$, thus applying Riesz's lemma to $\theta_n:=1-\frac1n$ and $Y\subset U$, we get a sequence $u_n$ with $||u_n||=1$ and $d(u_n,Y)>1-\frac1n$. Pick a convergent subseqence and let $z$ be its limit.

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You can make the prove even a little bit more compact: since the distance function $d(\cdot,Y) $ is continuous and the unit ball $B_U$ is compact in $U$, its image must be a compact subset of the real line. –  Elmar Zander Feb 15 '13 at 11:48
    
Well, as far as I know, the unit closed ball being compact is a sufficient condition for the space to be finite-dimensional. Can we also take it to be a necessary one, I wonder? –  Saaqib Mahmuud Feb 15 '13 at 20:08
    
In a finite dimensional space a set is compact if and only if it's closed and bounded. As the unit ball is closed and bounded, it's always compact in a finite dimensional space. –  Elmar Zander Feb 15 '13 at 21:42
    
Aha! You mean the closed unit ball. Well, now it makes sense. –  Saaqib Mahmuud Feb 16 '13 at 12:20
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