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$A$ be a subset of a Hilbert Space $H\ni \|x\|=1\forall x\in A$ and there is an $\epsilon\ni\|x-y\|\ge\epsilon\forall x,y\in A,x\neq y$ I need to know whether $A$ is finite or not.

Intuitively, It is clear that the set has not to be finite, the points satisfying these conditions are points on surface of a sphere.

Myshkin
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If $H$ is finite dimensional then $A$ must be finite - else keep adding points to $A$ to maintain the $\|x-y\|$ property until $A$ is maximal. Then the set of open balls of radius $\epsilon$ around elements of $A$ is an open cover of the unit sphere.

If $H$ is infinite-dimensional we should be able to use the Riesz Lemma to construct a counter example for any non-compact $H$:

Let $H$ be an infinite dimensional Hilbert space, and pick some $x \in H$ with $\|x\|=1$. Let $A = \{x\}$. Let $B= $span$(A)$. Then $B$ is closed, so by the Riesz Lemma, we can chose $x_1 \in H$ with $\|x_1\|=1$ and $\|x-x_1\|>\epsilon$. Add $x_1$ to $A$ and repeat - since $H$ is infinite dimensional, the process does not terminate.

Note that all of this would work in a regular normed space.

In a Hilbert space, explicitly, you can take any orthonormal basis - since $$\begin{align}\|x-y\|^2&=\|x\|^2 + \|y\|^2 = 2\end{align}$$by Pythagoras.

Mathmo123
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  • If $\dim H<\infty$, then unit ball will be compact and hence by your aargument we will not get any finite subcover , do you mean to say that? – Myshkin Oct 06 '14 at 10:34
  • The set of open balls of radius $\epsilon$ around elements of $A$ is an open cover that has no sub cover by construction. Hence $A$ must be finite – Mathmo123 Oct 06 '14 at 10:47
  • you are simply great, thanks for the answer, – Myshkin Oct 06 '14 at 10:49