Basic Analysis – Gently Done Topological Vector Spaces

16 Basic Analysis
Proposition 2.8 Let Nx be a neighbourhood base at a point x in a topological
space (X,T) and suppose that, for each U ∈ Nx , xU is a given point in U. Then
the net (xU ) Nx converges to x, where Nx is partially ordered by reverse inclusion.
Proof For any given neighbourhood W of x there is V ∈ N x such that V ⊆ W.
Let U � V. Then, by definition, U ⊆ V so that x U ∈ V. Thus x U ∈ W whenever
U � V and x U → x along N x .
We can characterize the closure of a set A as that set consisting of all the limits
of nets in A.
Theorem 2.9 Let A be a subset of a topological space (X,T). Then x ∈ A if and
only if there is a net (a α ) α∈I with a α ∈ A such that a α → x.
Proof We know that a point x ∈ X belongs to A if and only if every neighbourhood
of x meets A. Suppose then that (a α ) α∈I is a net in A such that a α → x.
By definition of convergence, (a α ) α∈I is eventually in every neighbourhood of x,
so certainly x ∈ A.
Suppose, on the other hand, that x ∈ A. Let V x be the collection of all
neighbourhoods of x ordered by reverse inclusion. Then V x is a directed set. We
know that for each U ∈ V x the set U ∩A is non-empty so let a U be any element
of U ∩A. Then a U → x.
The next result shows that sequences are sufficient provided that there are not
toomanyopensets, thatis, ifthespaceisfirstcountable. Toseethis, wefirstmake
an observation. Suppose that {A n : n ∈ N} is a countable neighbourhood base at
somegivenpointinatopologicalspace. Foreachn ∈ N, letB n = A 1 ∩A 2 ∩···∩A n .
Then {B n : n ∈ N} is also a neighbourhood base at the given point, but has the
additional property that it is nested; B n+1 ⊆ B n , for n ∈ N.
Theorem 2.10 Suppose that (X,T) is a first countable topological space and
let A ⊆ X. Then x ∈ A if and only if there is a sequence (a n ) n∈N in A such that
a n → x.
Proof The proof that a n → x implies that x ∈ A proceeds exactly as before.
For the converse, suppose that x ∈ A. We must exhibit a sequence in A which
converges to x. Since (X,T) is first countable, x has a countable neighbourhood
base {B n : n ∈ N}, say. By the remark above, we may assume, without loss of
generality, that B n+1 ⊆ B n for all n ∈ N. Since x ∈ A, B n ∩A �= ∅ for all n ∈ N.
So if we let a n be any element of B n ∩A, we obtain a sequence (a n ) n∈N in A. We
claim that a n → x as n → ∞. To see this, let G be any neighbourhood of x. Then
there is some member B N , say, of the neighbourhood base such that B N ⊆ G. If
n ≥ N then a n ∈ B n ∩ A ⊆ B N and so a n ∈ G and it follows that a n → x as
claimed.
Department of Mathematics King’s College, London