Download Compactness and total boundedness via nets The aim of this

Compactness and total boundedness via nets
The aim of this chapter is to define the notion of a net (generalized sequence) and
to characterize compactness and total boundedness by this important topological
notion.
Nets: definitions and main properties.
Definition 0.1. A pre-ordering on a nonempty set I is a binary relation ≤ on I
which is:
(a) reflexive (i.e., α ≤ α for each α ∈ I);
(b) transitive (i.e., if α ≤ β ≤ γ then α ≤ β).
The couple (I, ≤) will then be called a pre-ordered set.
Notice that the notion of a pre-ordered set is more general than the more common
notion of a partially ordered set: a partially ordered set is a pre-ordered set whose
pre-ordering is also antisymmetric (i.e., if α ≤ β and β ≤ α then α = β).
Definition 0.2. A directed set (or “upper-directed set”) is a pre-ordered set (I, ≤)
such that
∀α, β ∈ I ∃γ ∈ I : γ ≥ α, γ ≥ β.
For brevity, we shall write γ ≥ α, β to express γ ≥ α, γ ≥ β. Moreover, we shall
use the following quite useful terminology. Let P be a property of elements of a
directed set I. We shall say that:
• P holds eventually if there exists α0 ∈ I such that P holds for each α ≥ α0 ;
• P holds frequently if for each α ∈ I there exists β ≥ α satisfying P.
Thus “eventually” means “for all successors of some element”, and “frequently”
means “for arbitrarily large elements”.
Definition 0.3. Let E be a set. A net (or “generalized sequence”) in E is any
mapping
ϕ: I → E
where I is a directed set. Putting xα := ϕ(α) (α ∈ I), the net can be also denoted
by (xα )α∈I , or by (xα ).
Definition 0.4. Let (xα )α∈I be a net in a topological space X. We say that the
net converges to a point x ∈ X if each neighborhood U of x contains xα eventually
in I. In this case we write indifferently xα → x or limα xα = x .
Examples 0.5. Let X be a topological space.
(a) Each sequence in X is a net in X (with I = N), and its convergence as a
net coincides with its convergence as a sequence.
(b) Each function ϕ : R → X is a net in X (since R is a directed set). What is
the meaning of convergence for such a net?
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(c) The family U(x) of all neighborhoods of x is directed by the inverse inclusion
(U ≤ V ⇔ U ⊃ V ). Let ϕ : U(x) → X be such that ϕ(U ) ∈ U for every
U ∈ U(x). Then the net ϕ converges to x.
(d) Let f : [a, b] → R be a Riemann integrable function. The set P of all
partitions of [a, b] is directed by the inclusion. For every partition P =
{a = x0 < x1 < · · · < xn = b}, fix arbitrarily {ξ1 , . . . , ξn } ⊂ [a, b] such that
ξi ∈ [xi−1 , xi ] (1 ≤ i ≤ n), and consider the corresponding Riemannian sum
sP =
n
X
f (ξn )(xi − xi−1 ) .
i=1
Then {sP }P ∈P is a net in R that converges to
Rb
a
f (t) dt. (Verify the details!)
(e) Consider a three-point set I = {a, b, c} with a binary relation ≤ defined by
the following list:
a ≤ a, b ≤ b, c ≤ c, a ≤ b, b ≤ c.
Then I is a directed set and, if we put xa = 2, xb = −1, xc = 1, the net
(xγ )γ∈I converges (in R) to 1.
Now, we are ready for the definition of a subnet of a net.
Definition 0.6. Let (xα )α∈I be a net in a set E. A subnet of (xα ) is any net of the
form (xψ(β) )β∈J where J is a directed set and ψ : J → I is a mapping that “tends
to infinity” in the sense that for each α ∈ I, ψ(β) ≥ α eventually in J.
In other words, a subnet of a net ϕ : I → E is a composition ϕ◦ψ where ψ : J → I
is a mapping (from a directed set J) “tending to infinity”.
Example 0.7. Any subsequence of a sequence is also its subnet. However, a sequence can have subnets that are not subsequences. For instance, consider the sequence (n2 )n∈N and its subnet (m2 + 2mn + n2 )(m,n)∈N×N (where N × N is equipped
with the coordinate-wise partial order; what is the function ψ in this case?). Moreover, a subnet of a sequence need not be indexed by a countable set; for example,
the net ([t]2 )t∈R (where [t] denotes the lower integer part of t) is a subnet of the
sequence (n2 )n∈N .
Observation 0.8. Every subnet of a converging net (in a topological space) converges to the same limit.
Now, let us state a useful lemma about existence of certain subnets.
L:subnets
Lemma 0.9. Let E be a set, and (xα )α∈I a net in E. Let B be a family of subsets
of E, satisfying:
• xα is contained frequently in each element of B, and
• the intersection of any two elements of B contains an element of B.
Then (xα ) admits a subnet which is eventually contained in each element of B.
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Proof. Clearly, the family B is directed by the inverse inclusion. Consider the set
J = (α, B) ∈ I × B : xα ∈ B
equipped with the coordinate-wise pre-ordering. It is easy to see that J is a directed
set (verify this!). The function ψ : J → I, defined by ψ(α, B) = α, is “nondecreasing” and onto, an hence “tends to infinity”. Consequently, (xψ(α,B) )(α,B)∈J is a
subnet of (xα ). Moreover, given A ∈ B, fix α0 ∈ I so that xα0 ∈ A, and observe
that if (β, B) ≥ (α0 , A) then xψ(β,B) = xβ ∈ B ⊂ A. This completes the proof. Basic topological properties of nets. Let us list some basic results about nets
in topological spaces. We omit their relatively easy proofs; the interested reader can
find them in the book [J.L. Kelley, General Topology].
Theorem 0.10. Let X, Y be a topological spaces, A ⊂ X, x0 ∈ X.
(a) x0 ∈ A if and only if there is a net in A converging to x0 .
(b) A is closed if and only if there in no net in A converging to a point of X \A.
(c) X is Hausdorff if and only if each net converges to at most one point.
(d) A mapping f : X → Y is continuous at x0 if and only if it satisfies the
implication: X 3 xα → x0 ⇒ f (xα ) → f (x0 ) .
Compactness and nets. Now, we are going to prove a well-known characterization
of compactness via nets. The reader probably already knows its analogue for metric
spaces and sequences (instead of nets). The main difference between the two cases
is in their proofs.
Let us recall that a family A of sets is said to be centered if each finite subfamily
of A has nonempty intersection. It is an easy and well-known fact that a topological
space X is compact if and only if every centered family of closed subsets of X has
nonempty intersection.
cpt
Theorem 0.11. A topological space X is compact if and only if each net in X
admits a subnet converging to a point of X.
Proof. (⇒) Let (xα )α∈I be a net in a compact space X. Then the sets
Aα = {xβ : β ≥ α}
(α ∈ I)
form a centered family; hence theTfamily of their closures is centered as well. Since
X is compact, there exists x ∈ α∈I Aα . For each U ∈ U(x) (the family of all
neighborhoods of x) and each α ∈ I, one has U ∩ Aα 6= ∅, that is, xβ ∈ U for some
β ≥ α. But this means that the family U(x) satisfies the assumptions of Lemma 0.9.
Hence there exists a subnet of (xα ) that is eventually contained in each element of
U(x). In other words, this subnet converges to x.
(⇐) Assume that each net in X has a converging (in X) subnet. Let A be a
centered family of closed sets in X. Then also the family B, formed by all finite
intersections of elements of A, is centered, too. For each B ∈ B, choose an arbitrary
xB ∈ B. Since B is clearly directed by the inverse inclusion, (xB )B∈B is a net.
α
Let (xψ(α) )α∈I be a subnet of (xB ) such that xψ(α) → x ∈ X. Let us show that
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x belongs to each B ∈ B. Given B, we have ψ(α) ≥ B eventually in I, that is,
xψ(α) ∈ ψ(α) ⊂ B eventually in I. It follows that x = limα xψ(α) ∈ B = B, and the
proof is complete.
Total boundedness and nets. For defining the notion of total boundedness, we
need more specific class of spaces than merely topological spaces. The most general
appropriate structure is that of a uniform space, that includes metric spaces as well
as topological linear spaces as particular cases. However, uniform spaces are not
very common and working in such spaces is, in a sense, formally more difficult.
Moreover, the theorem, saying that a metric space E is totally bounded if and only
if each sequence in E has a Cauchy subsequence, is well-known and relatively easy,
due to fact that it is relatively easy to work with sequences. From these reasons,
here we confine ourselves to the case of subsets of a topological vector space. An
introduction to the theory of uniform spaces can be found in the book [J.L. Kelley,
General Topology].
Recall that a set E in a topological vector space is totally bounded (or precompact)
if for each U ∈ U(0) (the set of neighborhoods of 0) there is a finite set E0 ⊂ E such
that E ⊂ E0 + U .
It is easy to see that in normed spaces (or in topological metric spaces) this
definition coincides with the usual metric one: for each ε > 0 there is a finite set
E0 ⊂ E such that dist(x, E0 ) < ε for each x ∈ E.
Definition 0.12. Let X be a topological vector space.
(a) A net (xα )α∈I in X is said to be a Cauchy net if for each U ∈ U(0) there
exists α0 ∈ I such that xα − xβ ∈ U whenever α, β ≥ α0 .
(b) A set E ⊂ X is complete if each Cauchy net in E converges to a point of E.
tot.bdd
Theorem 0.13. Let E be a set in a topological vector space. Then E is totally
bounded if and only if each net in E admits a Cauchy subnet.
Proof. ⇒ Let (xα )α∈I be a net in a totally bounded set E. The family
A = {A ⊂ E : xα ∈ A frequently}
clearly contains E. Let B be a maximal subfamily of A that contains E and is closed
under making finite intersections (existence of such B follows by Zorn’s lemma). Let
us show several properties of B.
S
(a) If F is a finite subfamily of A such that F ∈ B, then F ∩ B =
6 ∅. Let
F = {A1 , . . . , An }. We claim that, for some index k, Ak ∩ B ∈ A for each B ∈ B.
Indeed, if this is not the case, for each i ∈ {1, . . . , n}
Tnthere exists
Sn Bi ∈ B such that
Ai ∩ Bi ∈
/ A; but then we get B 3 (A1 ∪ . . . ∪ An ) ∩ i=1 Bi ⊂ i=1 (Ai ∩ Bi ) ∈
/ A, a
contradiction. Our claim implies that the family of all finite intersections of elements
of B ∪ {Ak } is closed under finite intersections and is contained in A. By maximality
of B, we must have Ak ∈ B.
(b) For each set A ⊂ E, the family B contains either A or E \ A. If A ∈
/ A, then
eventually xα ∈ E \ A. Since the intersection of E \ A with any element of B belongs
to A, the family of all finite intersections of B ∪ {E \ A} is contained in A. Thus
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E \ A ∈ B by the maximality of B. In the same way we get that if E \ A ∈
/ A then
A ∈ B. Finally, if both A and E \ A belong to A then one of them belongs to B by
(a) (since E ∈ B).
(c) B contains arbitrarily small elements, in the sense that for each U ∈ U(0) there
exists B ∈ B such that B − B ⊂ U . Given U ∈ U(0), there exists V ∈ U(0) with
V − V ⊂ U . By total boundedness, there exists a finite set F := {y1 , . . . , yn } ⊂ E
such S
that E ⊂ F + V . Denoting Ei = (yi + V ) ∩ E (i = 1, . . . , n), we have
E = ni=1 Ei . Consider the set
S P = {i ∈ {1, . . . , n} : Ei ∈ A} and its complement
S
P c in {1, . . . , n}. Since C := i∈P c Ei ∈
/ A, we must have P 6= ∅. Let A = i∈P Ei .
Then E \ A ∈
/ A (since E \ A ⊂ C). By (b), we must have A ∈ B. By (a), there
exists k ∈ P with Ek ∈ B. Notice that Ek − Ek ⊂ V − V ⊂ U .
To conclude the proof of this implication, notice that the family B satisfies the
assumptions of Lemma 0.9. Hence there exists a subnet of (xα ) that is eventually
contained in each element of B. By (c), this subnet is Cauchy.
⇐ Assume that E is not totally bounded. There exists U ∈ U(0) such that
E \ (F + U ) 6= ∅ for each finite set F ⊂ U . An easy inductive construction gives a
sequence (xn ) in E such that xn+1 ∈
/ {x1 , . . . , xn } + U for each n. Since for each
two indexes m > n we have xm − xn ∈
/ U , our sequence has no Cauchy subnets. The
proof is complete.
The above theorem easily implies the following characterization of compactness.
cpt.tvs
Theorem 0.14. A set E in a topological vector space is compact if and only if E
is totally bounded and complete.
Proof. ⇒ Let E be compact. Given an open V ∈ U(0), the open cover {y + V :
y ∈ E} of E admits a finite subcover. This proves that E is totally bounded. Let
(xα ) be a Cauchy net in E. By Theorem 0.11, (xα ) admits a subnet converging to
a point of E. It easily follows that the net (xα ) converges to the same limit.
⇐ Assume E is totally bounded and complete. Given a net (xα ) in E, it admits
a Cauchy subnet by Theorem 0.13. Since E is complete, this subnet converges
to a point of E. Again, it follows that (xα ) converges to the same point. By
Theorem 0.11, E is compact.
Theorem 0.14 is an analogue of the well-known result about metric spaces: a metric space is compact if and only if it is totally bounded and (sequentially) complete.
Let us show that, in topological vector spaces with a countable base of neighborhoods of 0, sequential completeness and completeness coincide.
Recall that a family B ⊂ U(0) is a base of neighborhoods of 0 if each element of
U(0) contains some element of B.
Each t.v.s. which is metrizable clearly admits a countable base of U(0). It is a
well-known theorem that a topological vector space is metrizable if and only if it
is Hausdorff and admits a countable base of neighborhoods of 0 (and, in this case,
the metric can be chosen to be translation invariant) (see [W. Rudin, Functional
Analysis]). However, we shall not need this fact.
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Proposition 0.15. Let X be a topological vector space with a countable base of
neighborhoods of 0. A set E ⊂ X is complete if and only if E is sequentially
complete.
Proof. Let B = {Vn : n ∈ N} be a countable base of U(0). We can assume that
V1 ⊃ V2 ⊃ . . . ; indeed, otherwise we can substitute B with the base
{V1 , V1 ∩ V2 , V1 ∩ V2 ∩ V3 , . . .}.
⇒ Let E be complete, and (xn ) a Cauchy sequence in E. There exists a subnet
(xϕ(α) )α∈I converging to a point x ∈ E. Let us construct inductively a sequence
(αk ) in I. Choose α1 so that xϕ(α) ∈ x + V1 for each α ≥ α1 . If we already have
α1 , . . . , αk , choose αk+1 ≥ αk so that ϕ(αk+1 ) ≥ ϕ(αk ) + 1 and xϕ(α) ∈ x + Vk+1 for
each α ≥ αk+1 . It is easy to verify that (xϕ(αk ) )k∈N is a subsequence of (xn ) that
converges to x.
⇐ Let E be sequentially complete, and (xα )α∈I a Cauchy net in E. Let us
construct inductively a sequence (αk ) in I. Choose α1 so that xα − xα1 ∈ V1 for each
α ≥ α1 . If we already have α1 , . . . , αk , choose αk+1 ≥ αk so that xα − xαk+1 ∈ Vk+1
for each α ≥ αk+1 . Then (xαn ) is a Cauchy sequence since xαm −xαn ∈ Vn whenever
m ≥ n. Consequently, (xαn ) converges to a point x ∈ E. Now, it is easy to show
that (xα ) converges to x, too.