Basic Analysis – Gently Done Topological Vector Spaces
Basic Analysis – Gently Done Topological Vector Spaces
Basic Analysis – Gently Done Topological Vector Spaces
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18 <strong>Basic</strong> <strong>Analysis</strong><br />
Theorem 2.13 Let X and Y be topological spaces. A mapping f : X → Y is<br />
continuous if and only if whenever (x α ) I is a net in X convergent to x then the<br />
net (f(x α )) I converges to f(x).<br />
Proof Suppose that f : X → Y is continuous and suppose that (x α ) I is a net in<br />
X such that x α → x. We wish to show that (f(x α )) I converges to f(x). To see<br />
this, let V be any neighbourhood of f(x) in Y. Since f is continuous, there is a<br />
neighbourhood U of x such that f(U) ⊆ V. But since (x α ) converges to x, it is<br />
eventually in U, and so (f(x α )) I is eventually in V, that is, f(x α ) → f(x).<br />
Conversely, suppose that f(x α ) → f(x) whenever x α → x. Let V be open in<br />
Y. We must show that f −1 (V) is open in X. If this is not true, then there is<br />
a point x ∈ f −1 (V) which is not an interior point. This means that every open<br />
neighbourhood of x meets X \f −1 (V), the complement of f −1 (V). This is to say<br />
that x is a limit point of X \f −1 (V) and so there is a net (x α ) I in X \ f −1 (V)<br />
such that x α → x. By hypothesis, it follows that f(x α ) → f(x). In particular,<br />
(f(x α )) I is eventually in V, that is, (x α ) I is eventually in f −1 (V). However, this<br />
contradicts the fact that (x α ) I is a net in the complement of f −1 (V). We conclude<br />
that f −1 (V) is, indeed, open and therefore f is continuous.<br />
Nets can sometimes be useful in the study of the σ(X,F)-topology introduced<br />
earlier.<br />
Theorem 2.14 Let X be a non-empty set and let F be a collection of maps<br />
f λ : X → Y λ from X into topological spaces (Y λ ,S λ ), for λ ∈ Λ. A net (x α ) I in X<br />
converges to x ∈ X with respect to the σ(X,F)-topology if and only if (f λ (x α )) I<br />
converges to f λ (x) in (Y λ ,S λ ) for every λ ∈ Λ.<br />
Proof Suppose that x α → x with respect to the σ(X,F)-topology on X. Fix<br />
λ ∈ Λ. Let V be any neighbourhood of fλ (x) in Yλ . Then, by definition of the<br />
σ(X,F)-topology, f −1<br />
λ (V) is a neighbourhood of x in X. Hence (xα ) I is eventually<br />
in f −1<br />
λ (V), which implies that (fλ (xα )) I is eventually in V. Thus fλ (xα ) → fλ (x)<br />
along I, as required.<br />
Conversely, suppose that (xα ) I is a net in X such that fλ (xα ) → fλ (x), for<br />
each λ ∈ Λ. We want to show that (xα ) I converges to x with respect to the<br />
σ(X,F)-topology. Let U be any neighbourhood of x with respect to the σ(X,F)topology.<br />
Then there is m ∈ N, members λ1 ,...,λ m of Λ, and neighbourhoods<br />
Vλ1 ,...,V λm of fλ1 (x),...,f (x), respectively, such that<br />
λm<br />
x ∈ f −1<br />
λ1 (Vλ1 )∩···∩f−1 λm (V ) ⊆ U. λm<br />
By hypothesis, for each 1 ≤ j ≤ m, there is β j ∈ I such that f λj (x α ) ∈ V λj<br />
whenever α � β j . Let γ ∈ I be such that γ � β j for all 1 ≤ j ≤ m. Then<br />
xα ∈ f −1<br />
λ1 (Vλ1 )∩···∩f−1 λm (V ) ⊆ U λm<br />
whenever α � γ. It follows that xα → x with respect to the σ(X,F)-topology.<br />
Department of Mathematics King’s College, London