Re: little ask

In article <1173805140.131321.67690@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Antonio <anton.cali@xxxxxxxx> wrote:

Hi all,

Consider an infinite dimensional separable hilbert space H and denote
by S the following:

S := {x in H such that ||x||=1}.

Let K be a compact set in H.
Could someone help me to understand the reason for the intersection

S \cap K

is compact in S ?

Regards, Antonio


S is closed.
S \cap K is a closed subset of K.
Is that enough?
.

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