S orthonormal -> S countable

[Joubert]

Suppose X is a separable space endowed with an inner product. Is S is a
set of orthonormal vectors in X, it is finite or countably infinite.

The argument points out that if u,v belong to S then ||u-v||^2=2 so the
elements of S are at a certain positive distance from one another. It
then procedes to show that the number of spheres with centers in the
elements of S is at most equal to the number of elements of the
countable dense set that makes X separable, but I can't quite understand
how. Can anyone put it down for me?

Thanks in advance

[victor_me...@yahoo.co.uk]

For each u in S consider the ball B_u with centre u and radius 1/2.
Show that the B_u are pairwise disjoint. They form an uncountable
family of pairwise disjoint nonempty open sets. Consider a putative
countable dense set A in X. As A is countable it can't hit all of
the B_u; so there is some B_u containing no point of A ...

[smn]

For each u in S ,let u_ be the open ball centered at u of radius 1
half times square root of 2 .so that for different u ,the u_ 's are
disjoint . If x_j j=1,2,3... are the countable dense set then for each
u let j(u) be the smallest j with x_j in u_.(there is a j because the
x_j's are dense).
For different u ,the j(u) 's are different since the u_ 's are
disjoint .So u --> j(u) is a one -one map from S into the positive
integers .smn

[José Carlos Santos]

Let C be a countable dense subset of X. Each sphere (with radius 1)
centered at an element of S must contain some element of C. But the
intersection of any two such spheres is empty (unless it is twice the
same sphere, of course). If the set S was uncountable then since each
sphere contains some element of C and since there are as many spheres
as elements of S, C would be uncountable.

Best regards,

Jose Carlos Santos

[Joubert]

On 1/6/2010 12:20 AM, Jos� Carlos Santos wrote:

> Let C be a countable dense subset of X. Each sphere (with radius 1)
> centered at an element of S must contain some element of C. But the
> intersection of any two such spheres is empty (unless it is twice the
> same sphere, of course). If the set S was uncountable then since each
> sphere contains some element of C and since there are as many spheres
> as elements of S, C would be uncountable.
>
> Best regards,
>
> Jose Carlos Santos

Thanks, seems easy now.