Basic Analysis – Gently Done Topological Vector Spaces

14 Basic Analysis
ample, the problem seems to be that a point in a topological space may simply
have too many neighbourhoods to be penetrated by a sequence, or, put differently,
a sequence may not, in general, have enough elements to populate all the neighbourhoods
of the putative limit point. It is necessary to generalize the notion of a
sequence allowing an index set inherently more general than the natural numbers,
but retaining some concept of ‘eventually greater than’. The concept of directed
set is made for just this purpose. First we need the notion of a partial order.
Definition 2.2 A partially ordered set is a non-empty set P on which is defined a
relation � (called a partial ordering) satisfying:
(a) x � x, for all x ∈ P;
(b) if x � y and y � x, then x = y;
(c) if x � y and y � z, then x � z.
We sometimes write x � y to mean y � x.
Note that it can happen that a particular pair of elements of P are not comparable,
that is, neither x � y nor y � x need hold. In fact, a partial order on P
is more properly defined as a subset S, say, of the cartesian product P ×P, such
that:
(i) (x,x) ∈ S, for all x ∈ P;
(ii) if (x,y) and (y,x) belong to S, then x = y;
(iii) if (x,y) ∈ S and (y,z) ∈ S, then (x,z) ∈ S.
By writing x � y if and only if (x,y) ∈ S, we recover our original formulation
above.
Examples 2.3
1. Let P be the set of all subsets of a given set, and let � be given by set inclusion
⊆.
2. Let P = R, and take � to be the usual ordering ≤ on R.
3. Let P = R 2 , and define � according to the prescription (x ′ ,y ′ ) � (x ′′ ,y ′′ )
provided that both x ′ ≤ x ′′ and y ′ ≤ y ′′ in R.
4. Any subset of a partially ordered set inherits the partial ordering and so is itself
a partially ordered set.
Definition 2.4 A directed set is a partially ordered set I, with partial order �, say,
such that for any pair of elements α,β ∈ I there is some γ ∈ I such that α � γ
and β � γ.
It follows, by induction, that if x 1 ,...,x n is any finite number of elements of
a directed set I, there is some z ∈ I such that x i � z for all i = 1,...,n.
Examples 2.5
1. Let I = N (or Z or R) furnished with the usual order.
Department of Mathematics King’s College, London