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Basic Analysis – Gently Done Topological Vector Spaces

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120 <strong>Basic</strong> <strong>Analysis</strong><br />

because X = �∞ k=1kV2 , since V2 is a neighbourhood of 0 and so is absorbing. By<br />

Baire’s category theorem, k0T(V2 ) has non-empty interior for some k0 ∈ N. But<br />

y ↦→ k0y is a homeomorphism of Y onto Y so that k0T(V2 ) = k0T(V2 ) and we<br />

deduce that T(V2 ) has non-empty interior, as claimed.<br />

In particular, this means that there is an open set U contained in T(V2 ).<br />

Putting W = U −U, we see that W is an open neighbourhood of 0 with<br />

W ⊆ T(V 2 )−T(V 2 ) ⊆ T(V 1 )<br />

which completes the first part of the argument.<br />

Next we shall show that T(V 1 ) ⊆ T(V). To see this, we construct a sequence<br />

(y n ) n∈N , with y n ∈ T(V n ), such that y n+1 − y n ∈ T(V n ). The y n are defined<br />

recursively. First fix y 1 to be any point of T(V 1 ). Suppose that n ≥ 1 and that y n<br />

has been chosen in T(V n ). Arguing as above, but now with V n+1 instead of V 1 , we<br />

see that T(V n+1 ) is a neighbourhood of 0. Hence y n −T(V n+1 ) is a neighbourhood<br />

of y n , and, since y n belongs to the closure of T(V n ), we have<br />

� yn −T(V n+1 ) � ∩T(V n ) �= ∅.<br />

Hence there is x n ∈ V n such that T(x n ) ∈ y n −T(V n+1 ). Set y n+1 = y n −T(x n ).<br />

Then y n+1 ∈ T(V n+1 ) which completes the construction of the sequence (y n ).<br />

Let s n = x 1 +x 2 +···+x n , for n ∈ N. Then for n > m,<br />

d(s n ,s m ) = d(s n −s m ,0)<br />

= d(x n +···+x m+1 ,0)<br />

≤ d(xn +···+x m+1 ,xm+1 )+d(x m+1 ,0)<br />

< d(xn +···+x m+2 ,0)+ 1<br />

2m+1 since xm+1 ∈ Vm+1 ,<br />

< 1 1<br />

+···+<br />

2n 2m+1. Hence (s n ) is a Cauchy sequence in X and therefore converges (because X is<br />

complete) to some x ∈ X, with d(x,0) < r (since d(x,0) ≤ d(x,x 1 )+d(x 1 ,0) <<br />

d(x,x 1 ) + r/2 and d(s n ,x 1 ) = d(x n + ··· + x 2 ,0) ≤ r/2 for all n > 1, so that<br />

d(x,x 1 ) = limd(s n ,x 1 ) ≤ r/2). It follows that x ∈ V 0 ⊆ V. Furthermore,<br />

T(s n ) =<br />

n�<br />

T(xj ) =<br />

j=1<br />

n�<br />

j=1<br />

(y j −y j+1 ) = y 1 −y n+1 .<br />

Now, y n ∈ T(V n ) and so there is some v n ∈ V n such that d(y n ,T(v n )) < 1<br />

n and<br />

hence<br />

d(y n ,0) ≤ d(y n ,T(v n ))+d(T(v n ),0) < 1<br />

n +d(T(v n ),0).<br />

Department of Mathematics King’s College, London

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