Question #133627
Show that Hilbert space is not locally compact at any point.
1
Expert's answer
2020-09-17T17:05:43-0400
  • A topological space is called locally compact if, roughly speaking, each small portion of the space looks like a small portion of a compact space.A space Xis locally compact at a point x in X provided that there is an open set U containing x for which U  is compact. X is locally compact provided that it is locally compact at each point.

Proof:

Seeking to prove Hilbert Space H is not locally compact at any point by contradiction. Suppose H

 is locally compact at a point p=(x1,x2,....)Let U\overline{U}  be an open set containing p. Since H

H is locally compact . U\overline{U} is compact; thus r>0:B(p,r)U\exists r>0:B(p,r)\subset U, then B(p,r)=B(p,r)U\overline{B(p,r)}=B(p,r)\subset \overline{U}

However, the set P= {pn} n=1 to infinity of points pn=(x1,x2,...,xn1,xn+r/2,xn+1,...)p_n= (x_1,x_2,...,x_{n-1},x_n+r/2,x_{n+1},...) is an infinite subset ofB[p,r]B[p,r] with no limit point. Since compactness is equivalent to the Bolzano-Weierstrass property in metric spaces, we must conclude that B[p,r]B[p,r] is not compact. Thus U\overline{U} is not compact and H is not locally compact at any point.



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Assignment Expert
18.09.20, 20:22

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PRATHIBHA ROSE C S
18.09.20, 07:01

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