In Chapter 1, we defined limits of sequences in a topological space (X, T) so as to assimilate the metric definition. We noted, however, that not everything we knew about this idea in metric spaces is valid in topological spaces.
We will examine two main ways around this difficulty:
The following important results are probably familiar to us in the context of metric spaces, or at least in the setting of the real line, R.
Theorem 1 Given (X,T), A Í X, p Î X: if there exists some sequence of points of A tending to p, then p Î [`A].
Theorem 2 Given (X,T), A Í X: if A is closed, then A includes the limit of every convergent sequence of points of A.
Theorem 3 Given f:(X,T) ® (Y,T¢): if f is continuous, then f `preserves limits of sequences' i.e. whenever xn ® l in X, then f(xn) ® f(l) in Y.
In each case above, it is routine to prove the statement true in a general topological space as asserted. We illustrate by proving Theorem 3.1:
Let f be continuous and xn ® l in X. We must show that f(xn) ® f(l). Given a neighbourhood N of f(l), there exists open G such that f(l) Î G Í N. Then l Î f-1(G) Í f-1(N) i.e. f-1(N) is a neighbourhood of l and so xn Î f-1(N) "n ³ n0 say. Thus f(xn) Î N "n ³ n0, whence f(xn) ® f(l).
In metric spaces, the converses of these results are also true but our main point here is that in general topology, the converses are not valid.
Example
In (R, L), [`(0,1)] = R. So, for example, 5 Î [`(0,1)] and yet the only way a sequence (xn) converges to a limit
l is for xn = l from some stage on. So no sequence in (0,1)
can converge to 5 proving that the converse of Theorem 3.1 is false.
Continuing, the limit of any convergent sequence in (0,1) must belong to (0,1) for the same reason and yet (0,1) is not closed. Thus, Theorem 3.1's converse is false.
Further, idR:(R, L) ® (R, Á) is not continuous and yet it does preserve limits of sequences.
Now this is a great nuisance! Sequences are of immense usefulness in real analysis and in metric spaces and elsewhere - and their failure to describe general topology adequately is a technical handicap. What to do?
Recall that a sequence is just a function having the positive integers as domain. The set of positive integers, of course, possesses a particularly simple ordering; there is a first member, second member, third member, etc. Not all sets are naturally endowed with so simple an ordering. For example, dictionary (lexographical) ordering of words is more complex (though still relative nice as orderings go). By replacing the domain of positive integers with a set having a more complicated ordering we will:
Note that these last two items generalize the role of sequences in a metric space.
Definition 1
A binary relation £ on a set P is said to be a pre-order iff
(i) p £ p "p Î P
(ii) p £ q and q £ r imply p £ r "p,q,r Î P.
If it is also true that for p,q Î P,
{\em (iii)} p £ q and q £ p imply p = q, P is said to be a
partially ordered set (or poset).
Definition 2 A pre-ordered set P is said to be directed (or updirected) iff each pair of members of P has an upperbound.
(i.e. if p,q Î P, then there exists s Î P such that p £ s, q £ s.)
Definition 3
Let (P, £ ) be a poset. Then if x,y Î P with x \not £ y and
y \not £ x, we write x ||y and say that x and y are
incomparable;
If E Í P, then E is said to be totally unordered (or diverse)
iff x,y Î E implies x = y or x ||y.
If C Í P, then C is said to be linear (or a chain or
a total order) iff x,y Î C implies x < y, x = y or y < x.
(P, £ ) is said to be a lattice iff each pair of members of P
has a greatest lower bound and a least upper bound.
A lattice (P, £ ) is said to be complete iff every non-empty
subset Y of P has a greatest lower bound (ÙY) and a least upper
bound (ÚY).
An element v of a poset (P, £ ) is said to be maximal
( minimal) iff v £ x (x £ v), x Î P Þ v = x.
Definition 4 A net in a (non-empty set) X is any function x:A ® X whose domain A is a directed set.
In imitation of the familiar notation in sequences, we usually write the net value x(a) as xa. A typical net x:A ® X will usually appear as (xa, a Î A) or (xa)a Î A or some such notation.
(i)
N, Z, N ×N are all directed sets, where suitable
pre-orders are respectively the usual magnitude ordering for N and Z,
and (i,j) £ (m,n) iff i £ m and j £ n, in N ×N.
Thus, for example, a sequence is an example of a net.
(ii)
The real function f:R \{0} ® R given by
f(x) = 3-[1/x] is a net, since its domain is a chain. Any
real function is a net.
(iii)
Given x Î (X, T), select in any fashion an element
xN from each neighbourhood N of x; then (xN)N Î Nx
is a net in X ( since it defines a mapping from (Nx, £ )
into X). Recall that Nx is ordered by inverse set inclusion!
Definition 5 A net (xa)a Î A in (X, T) converges to a limit l if for each neighbourhood N of l, there exists some aN Î A such that xa Î N for all a ³ aN.
In such a case, we sometimes say that the net (xa)a Î A
eventuates N. Clearly, this definition incorporates the old definition of `limit of a sequence'. The limit of the net f described in (ii) above is 3. In (iii), the net described converges to x no matter how the values xN are chosen ... prove!
Our claim is that nets `fully describe' the structure of a topological space. Our first piece of evidence to support this is that with nets, instead of sequences, Theorems 3.1, 3.1 and 3.1 have workable converses:
Theorem 4 Given (X,T), A Í X, p Î X: p Î [`A] iff there exists a net in A converging to p.
Proof If some net of points of A converges to p, then every neighbourhood of p contains points of A (namely, values of the net) and so we get p Î [`A]. Conversely, if p is a closure point of A then, for each neighbourhood N of p, it will be possible to choose an element aN of A that belongs also to N. The net which these choices constitute converges to p, as required.
Theorem 5 Given (X,T), A Í X, A is closed iff it contains every limit of every (convergent) net of its own points.
Proof This is really just a corollary of the preceding theorem.
Theorem 6 Given f:(X,T) ® (Y,T¢), f is continuous iff f preserves net convergence.
Proof Exercise.
Definition 6 Let (xa)a Î A be any net and let a0 Î A. The a0th tail of the net is the set {xa: a ³ a0} = x([a0, )). We denote it by x(a0 ®).
Definition 7
Let (xa)a Î A and (yb)b Î B be any two nets.
We call (yb)b Î B a subnet of (xa)a Î A
provided that every tail of (xa) contains a tail of (yb) i.e.
provided:
"a0 Î A $b0 Î B such that x(a0 ®) Ê y(b0 ®).
We expected a definition like `subsequence' to turn up here and we are disappointed that it has to be so complicated.
Net theory ceases to be a straightforward generalisation of sequence theory precisely when we have to take a subnet ... so we'll try to avoid this whenever possible! There is however one result certainly worth knowing:
Theorem 7 (X, T) is compact iff in X, every net has (at least one) convergent subnet.
(So, for example, (n) is a net in R with no convergent subnet.)
Proof Not required.
Corollary 1 Compactness is closed-hereditary
Proof (for if (xa) is a net in a closed set F Í X, then it has a convergent subnet (yb) in X. Thus there exists a subnet (zg) of (yb) in F which converges in X, whence its limit is in F).
Corollary 2 Compactness is preserved by continuous maps
Proof (for if X is compact and f continuous, let (ya)a Î A be a net in f(X). Then for each a Î A, ya = f(xa) for some xa Î X. The net (xa)a Î A has a convergent subnet (zb)b Î B, say zb ® l, whence f(zb) ® f(l). Then (f(zb))b Î B is a convergent subnet of (ya)a Î A).
Example
If (xnk) is a subsequence of a sequence (xn), then it is a subnet of
it also; because the i0th tail of the sequence (xn) is
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Lemma 1 If a net (xa) converges to a limit l, then so do all its subnets.
Proof Let (yb) be a subnet of (xa); let N be a neighbourhood of l. Then there exists a0 such that xa Î N for all a ³ a0. Further, there exists b0 such that {yb: b ³ b0} Í {xa:a ³ a0} and so yb Î N for all b ³ b0.
Definition 8 Let x Î (X, T). A countable neighbourhood base at x means: a sequence N1, N2, N3, ... of particular neighbourhoods of x such that every neighbourhood of x shall contain one of the Ni's.
Note that we may assume that N1 Ê N2 Ê N3 Ê ¼ because, if not, then we can work with N1, N1 ÇN2, N1 ÇN2 ÇN3, ...
Definition 9 We call (X, T) first-countable when every point in X has a countable neighbourhood base.
Example The classic example of a first-countable space is any metric (or metrizable) space because if x Î (M,d), then B(x,1), B(x, [1/2]), B(x, [1/3]), ... is a countable neighbourhood base at x.
Theorem 8 First-countability is hereditary and preserved by continuous open onto maps.
Proof Left to the reader.
Theorem 9
(i) Complete separability implies first countability.
(ii) Converse not always true.
(iii) Converse valid on a countable underlying set.
(i) If B is a countable base for (X, T) and p Î X,
consider { B Î B: p Î B} which is a countable family of
neighbourhoods of p. Moreover, they form a neighbourhood base at p.
(ii) An uncountable discrete space is first countable ( since metrizable),
yet is not completely separable.
(iii) Suppose X countable and (X, T) first countable. For
each x Î X, choose a countable neighbourhood base: N(x,1), N(x,2),
N(x,3), .... Each is a neighbourhood of x and so contains an open
neighbourhood of x: G(x,1), G(x,2), G(x,3), ....
Then B = { G(x,n) :n Î N, x Î X} is a countable family of open sets and is a base for (X, T). Thus, (X, T) is completely separable.
Example
The Arens-Fort space (see, for example, Steen and Seebach, Counterexamples
in Topology is not first-countable because otherwise it would be
completely separable which is false!
Theorem 10
Given a first-countable space (X, T)
(i) p Î X, A Í X, then p Î [`A] iff there exists
a sequence of points of A converging to p.
(ii) A Í X is closed iff A contains every limit of every
convergent sequence of its own points.
(iii) f:(X,T) ® (Y,T¢) is continuous iff
it preserves limits of (convergent) sequences.
(i) Theorem 3.1 said that if there exists a sequence in A converging to some
p Î X, then p Î [`A].
Conversely, if p Î [`A], then p has a countable base of
neighbourhoods N1 Ê N2 Ê N3 Ê ¼, each of
which must intersect A. So choose xj Î Nj ÇA for all j ³ 1.
Then (xj) is a sequence in A and, given any neighbourhood H of p,
H must contain one of the Nj's i.e. H Ê Nj0 Ê Nj0 + 1 Ê ¼ so that xj Î H for all j ³ j0.
That is, xj ® p.
(ii) Corollary of (i).
(iii) f continuous implies that it must preserve limits of sequences
(by Theorem 3.1). Conversely, if f is not continuous, there exists A Í X such that f([`A]) \not Í [`f(A)]. Thus,
there exists p Î f([`A]) \[`f(A)] so p = f(x), some
x Î [`A]. So there exists a sequence (xn) in A with xn ® x.
Yet, if f(xn) ® f(x) ( = p), p would be the limit of a sequence in f(A) so that p Î [`f(A)] -contradiction! Thus f fails to preserve convergence of this sequence.