NETS IN TOPOLOGICAL SPACES – INADEQUACY OF SEQUENCES to ADEQUACY OF NETS

 

NETS IN TOPOLOGICAL SPACES

– INADEQUACY OF SEQUENCES to ADEQUACY OF NETS

M. VELRAJAN

In a metric space, sequences are adequate for closed sets, continuous functions and

compact spaces. 

i.e. in a metric space closed sets, continuous functions and compact spaces are

determined by sequences. 

In fact,

1. Suppose A is a nonempty subset of a metric space X and  A be the closure of A

in X.

Then x ∈ A if and only if there exists a sequence (an) in A such that an → x in X.

2. Suppose (X, dX) and (Y, dY) are metric spaces and f : X → Y is a function.

Then f is continuous at a ∈ X if and only if for all sequences an → a in X, the 

sequence f(an) → f(a) in Y. 

3. A metric space X is compact if and only if X is sequentially compact 

i.e. every sequence in X has a convergent subsequence. 

But in topological space only one part of 1. and 2. are true but the other part is not true, and both parts of 3. are not true.

1′. Suppose A is a nonempty subset of a topological space X.

If there exists a sequence (an) in A such that an → x in X 

then x ∈ A, the closure of A. 

But if x ∈ A then there need not exists a sequence (an) in A

such that an → x in X. (Think and get an example)

2′. Suppose X and Y are topological spaces, f : X → Y be a function and a ∈ X.

If f is continuous at a then for all sequences an → a in X,  f(an) → f(a) in Y.

But if for all sequences an → a in X,  f(an) → f(a) in Y then f need not be

continuous at a. (Think and get an example)

3′.  There are compact topological spaces which are not sequentially compact 

and there are sequentially compact topological spaces which are not compact.

 

So the natural question arises within us is :

Why do the other parts of 1′ and 2′ are true in metric spaces and not true in

topological spaces ? 

To get an answer for this question we go through the proof of 1. and 2. 

Usual proof of them are given below :  

1. Suppose A is a nonempty subset of a metric space X.

Then x ∈ A if and only if there exists a sequence (an) in A such that an → x in X.

And x is a limit point of A if and only if there exists a sequence (an) of distinct elements

of A such that an → x in X. 

Proof   Since A = A ∪ D(A), where D(A) is the set of all limit points of A. 

x ∈ A  if and only if x ∈ A or x is a limit point of A.

If x ∈ A then the constant sequence x, x, x, . . . in A converges to x in X.

If x  is a limit point of A then for each r > 0, the open ball B(x, r) contains

infinitely many points of A. In particular,

for each n ∈ ℕ, the open ball B(x, 1n) contains infinitely many points of A.

Choose a1 ∈ A ∩ B(x, 1). 

Since A ∩ B(x, 1/2) contains infinitely many points of A, we can choose

a2 ∈ A ∩ B(x, 1/2) such that a2 ≠ a1.

If a1, a2, . . . , an are chosen such that ai ∈ A ∩ B(x, 1/i) and

ai ≠ aj, j = 1, 2, . . . , (i - 1), i = 1, 2, . . . , n then 

since A ∩ B(x, 1/(n + 1) contains infinitely many points of A, we can choose

an + 1 ∈ A ∩ B(x, 1/(n + 1) such that an + 1 ≠ ai, i = 1, 2, . . . , n.

Hence by induction, we get a sequence (an) of distinct elements of A such that 

an ∈ A ∩ B(x, 1/n) i.e. such that an ∈ A and an ∈ B(x, 1/n), for all n.

Since 1/k ≤ 1/n for all k ≥ n, ak ∈ B(x, 1/k) ⊆ B(x, 1/n), for all k ≥ n

Let ϵ > 0. Since the sequence 1/n → 0, there exists n0 such that 1/n < ϵ, for all n ≥ n0.

Hence ak ∈ B(x, 1/k) ⊆ B(x, 1/n0) ⊆ B(x, ϵ), for all k ≥ n0.

Hence an → x in X.

Conversely, if there exists a sequence (an) in A such that an → x in X then 

for each ϵ > 0 there exists a natural number N such that d(an, x) < ϵ  for all n ≥ N 

i.e. an ∈ {y ∈ X / d(y, x) < ϵ} = B(x, ϵ), for all n ≥ N. 

Since an ∈ A, every open ball B(x, ϵ) contains a point of A.

Hence x ∈ A.

If there exists a sequence (an) of distinct elements in A such that an → x in X then

for each ϵ > 0 there exists a natural number N such that d(an, x) < ϵ  for all n ≥ N

i.e. an ∈ {y∈ X / d(y, x) < ϵ} = B(x, ϵ), for all n ≥ N. 

Since an ∈ A are distinct, every open ball B(x, ϵ) contains infinitely many 

points of A. Hence x is a limit point of A.

2.  Suppose (X, dX) and (Y, dY) are metric spaces and f : X → Y be a function.

Then f is continuous at a ∈ X if and only if for all sequences an → a in X, the

sequence f(an) → f(a) in Y. 

Proof      Suppose f is continuous at a ∈ X.

Suppose the sequence an → a in X. Let 𝛜 > 0. Since f is continuous at a, 

∃ 𝛅 > 0 such that if x ∈ X and dX(x, a) < 𝛅 then dY(f(x), f(a)) < 𝛜.

Since an → a in X, for the 𝛅 > 0, ∃ m such that dX(an, a) < 𝛅, for all n ≥ m.

Hence dY(f(an), f(a)) < 𝛜, for all n ≥ m.

So, the sequence f(an) → f(a) in Y. 

Conversely, suppose for all sequences an → a in X, the sequence f(an) → f(a) in Y. 

(We cannot proceed from the given hypothesis, so we prove contrapositively.)

Suppose f is not continuous at a.

Then there exists ϵ > 0 such that, for all δ > 0, there exists 

x ∈ X with dX(x, a) < δ but dY(f(x), f(a)) ≥ ϵ.

(Note how to write negation, by replacing ‘for all’ by ‘there exists’,

‘there exists’ by ‘for all’, ‘<’ by ‘≥’, etc)  

In particular, for each n ∈ ℕ, there exists 

xn ∈ X with dX(xn, a) < 1n i.e. xn ∈ B(a, 1/n),  but dY(f(xn), f(a)) ≥ ϵ. 

Since the sequence 1/n → 0, 

for the ϵ > 0, ∃ m such that |1/n - 0| = 1/n < 𝛜, for all n ≥ m. 

Hence dX(xn, a) < 1/n < 𝛜, for all n ≥ m. So, xn → a in X.

But since for each n ∈ ℕ, dY(f(xn), f(a)) ≥ ϵ, f(xn) ⇸ f(a).

Which contradicts the hypothesis, 

for all sequences an → a in X, f(an) → f(a) in Y.

Hence f is continuous at a. 

 

After going through the above proofs we can observe a common thing in the

proof of the other parts of 1 and 2. What is that?

Yes, both of them use the countable open balls B(a, 1/n), n = 1, 2, . . . . and the

fact that B(a, 1/m) ⊆ B(a, 1/n), for all m ≥ n.

And of course for any open set U containing a, there exists n such that

B(a, 1/n) ⊆ U. 

Also we observe from the proofs that : 

4.  In a metric space if an ∈ B(a, 1/n), the open ball with centre a and 

radius 1/n, for all n ∈ ℕ then an → a.

Recall : 

5.  In a metric space a sequence an → a if for every ϵ > 0 there exists a natural

number m such that an ∈ B(a, ϵ), for all n ≥ m, where B(a, ϵ) is the open ball 

with centre a and radius ϵ. 

i.e. an → a if for every ϵ > 0, B(a, ϵ) contains all terms of (an) except

a finite number of terms.

From our observation 4., in the above definition   

‘The uncountable set {B(a, ϵ) / ϵ > 0}’ can be replaced by ‘its countable subset

{B(a, 1/n) / n ∈ ℕ}’ and ‘B(a, ϵ) contains all terms of (an) except a finite

number of terms’ by ‘an ∈ B(a, 1/n) for all n ∈ ℕ’.

Now we think about how to extend this to topological spaces.

Recall

6.  Suppose (an) is a sequence in a topological space X, a ∈ X and 𝓑 is a basis at a.

Then an → a in X if and only if for each B ∈ 𝓑 there exists m ∈ ℕ such that 

an ∈ B, for all n ≥ m.

Of course in general topological space instead of open balls B(a, ϵ) about a of a

metric space, we have the counterpart basis elements at a.  

And for any open set U containing a, there exists a basis element B at a 

such that B ⊆ U.

And in general the basis at a is not countable. But,

7. In a first countable topological space X, every a ∈ X has a countable basis 

{Bn / n ∈ ℕ} at a. 

But Bn need not contain Bn+1, for all n ≥ 1.

Think how to get a decreasing countable basis at a.

Yes, 

8. If {Bn / n ∈ ℕ} is a countable basis at a and 

Cn = B1 ∩ B2 ∩ . . . ∩ Bn, for all n ≥ 1 then 

{Cn / n ∈ ℕ} is also a countable basis at a and Cn ⊇ Cn+1, for all n ≥ 1. (Verify !)

i.e. In a first countable topological space X, every a ∈ X has a countable basis 

{Cn / n ∈ ℕ} at a such that Cn ⊇ Cn+1, for all n ≥ 1.

So, in a first countable topological space Cn are equivalent to B(a, 1/n)

in metric space. 

And equivalent to 4 we have :

9. Suppose (an) is a sequence in a first countable topological space X, a ∈ X and

{Bn / n ∈ ℕ} is a countable basis at a and Cn = B1 ∩ B2 ∩ . . . ∩ Bn, for all n ≥ 1.

If an ∈ Cn, for all n ∈ ℕ then an → a in X. 

For,

By 8., {Cn / n ∈ ℕ} is also a countable basis at a and Cn ⊇ Cn+1, for all n ≥ 1.

Let n ∈ ℕ. Since Cn ⊇ Cn+1, for all n ≥ 1 and ak ∈ Ck, for all k ∈ ℕ we have

ak ∈ Cn, for all k ≥ n. Hence, By 6., an → a in X. 

Hence, for a first countable topological space X, we can replace B(a, 1/n) by Cn

in the proof of the other parts of 1′ and 2′.

So, 1′ becomes

10. Suppose A is a nonempty subset of a topological space X.

If there exists a sequence (an) in A such that an → x in X 

then x ∈ A, the closure of A. 

And the converse holds if X is first countable. 

Proof   If there exists a sequence (an) in A such that an → x in X then 

for each neighbourhood V of x there exists a natural number N such that 

an ∈ V for all n ≥ N. 

Since an ∈ A, every neighbourhood V of x contains a point of A.

Hence x ∈ A.

Conversely, suppose X is first countable and x ∈ A.

Suppose {Bn / n ∈ ℕ} is a countable basis at x and 

Cn = B1 ∩ B2 ∩ . . . ∩ Bn, for all n ≥ 1.

Then A ∩ Cn is non-empty, for all n ≥ 1.

Let an ∈ A ∩ Cn, for all n ≥ 1. Then an ∈ A for all n ≥ 1 and an ∈ Cn, for all n ≥ 1. 

By 9., an → x in X. 

Recall : Suppose X satisfies T1 axiom : Every finite subset of X is closed in X.

And A is a subset of X. Then x ∈ X is a limit point of A if and only if every

neighbourhood of x contains infinitely many points of A.

10.1.  Suppose X is first countable and satisfies T1 axiom. 

And A is a subset of X. Then x ∈ X is a limit point of A if and only if there exists a

sequence (an) of distinct elements of A such that an → x in X. 

And 2′ becomes

11. Suppose X and Y are topological spaces, f : X → Y be a function and a ∈ X.

If f is continuous at a then for all sequences an → a in X,  f(an) → f(a) in Y.

And the converse holds if X is first countable.

Proof  Suppose f is continuous at a ∈ X.

Suppose the sequence an → a in X. Let V be any open set in Y containing f(a). Since f is continuous at a, f – 1(V) is open in X and a ∈ f – 1(V). 

Since an → a in X,  ∃ m such that an ∈ f – 1(V) for all n ≥ m.

Hence f(an) ∈ V for all n ≥ m.

So, the sequence f(an) → f(a) in Y. 

Conversely, suppose for all sequences an → a in X, the sequence f(an) → f(a) in Y. 

Suppose {Bn / n ∈ ℕ} is a countable basis at a and 

Cn = B1 ∩ B2 ∩ . . . ∩ Bn, for all n ≥ 1.

(We cannot proceed from the given hypothesis, so we prove contrapositively.)

Suppose f is not continuous at a.

Then there exists an open set V in Y such that f(a) ∈ V and f - 1(V) is not open in X.

(Note how negation is written, by replacing ‘for all’ by ‘there exists’, 

‘open in X’ by ‘not open in X’)  

Since a ∈ f - 1(V) and {Cn / n ∈ ℕ} is a basis at a, Cn ⊈ f - 1(V) for all n ≥ 1.

(if Cn ⊆ f - 1(V) for some n ≥ 1 then f - 1(V) is open in X)

Hence for each n ≥ 1, there exists xn ∈ Cn such that xn ∉ f - 1(V) 

Since xn ∈ Cn, for all n ≥ 1,  By 9., xn → a in X.

Since f(xn) ∉ V, for all n ≥ 1 and V is a neighbourhood of f(a), f(xn) ⇸ f(a).

Which contradicts the hypothesis, 

for all sequences an → a in X,  f(an) → f(a) in Y.

Hence f is continuous at a. 

So a deep observation of proofs of 1 and 2 gives answers to the questions that

arise within us.

And we observe that in first countable topological spaces, sequences are adequate

to the topological properties : closed sets and continuous functions, as in the case

of metric spaces.

Instead of imposing conditions on topological space, the other way is thinking of

generalisation of sequences.

In 10. we observe that X is first countable i.e. the countable basis at x is used 

in the proof of : “if x ∈ A then there exists a sequence (an) in A such that 

an → x in X” only to get the countable collection of points of A, 

i.e. the sequence (an) in A.

So to avoid the first countable condition, we can generalise a sequence 

{an / n ∈ ℕ}, i.e. a function from ℕ into X, 

by a collection of points {aα / α ∈ J}, i.e. a function from J into X, 

where J is a set which generalise the set ℕ of all natural numbers, but need not

be a countable set.

 

Since ℕ is an ordered set with usual ≤, J has to be a partially ordered set. 

In the proof of sequence properties we use the property of natural numbers

that for each m, n ∈ ℕ, max{m, n} ∈ ℕ, hence m ≤ max{m, n} and 

n ≤ max{m, n}. 

A partially ordered set need not have this property, but a directed set has this

property. 

12.  A directed set J is a set with a partial order ⪯ such that for each α, β ∈ J,

there exists an element 𝛾 ∈ J such that α ⪯ 𝛾 and β ⪯ 𝛾. 

13.  (a) ℕ is a directed set, under the usual relation ≤. 

Verify that the following are directed sets :

(b) The collection of all subsets of a set S, partially ordered by inclusion 

(that is, A ⪯ B if A ⊆ B). 

(c) A collection A of subsets of S that is closed under finite intersections, partially

ordered by reverse inclusion (that is A ⪯ B if A ⊇ B). 

(d) The collection of all closed subsets of a space X, partially ordered by inclusion. 

14.  Let X be a topological space. A net in X is a function f from a directed set J

into X. If α ∈ J, we usually denote f(α) by xα. We denote the net f by the symbol (xα)α ∈ J, or merely by (xα) if the index set is understood. 

The net (xα) is said to converge to the point x of X (written xα → x) if for each

neighborhood U of x, there exists α ∈ J such that α ⪯ β ⇒ xβ ∈ U. 

Observe that a net reduce to sequence when J = ℕ.

15.  If X is Hausdorff, a net in X converges to at most one point. (Verify!)

16.  Let X be a topological space and x ∈ X. 

Let J be the collection of all neighborhoods of x.

Suppose for each U ∈ J, aU ∈ U.

We denote {aU / U ∈ J} by (aU)U ∈ J then (aU)U ∈ J is a net in A and aU → x.

For,  Since J is closed under finite intersections, by 13(c), 

J is a directed set by the reverse inclusion ⊇.

Hence (aU)U ∈ J is a net in A.

For each neighborhood U of x, 

U ∈ J and for any V ∈ J such that U ⊇ V, aV ∈ V ⊆ U.

Hence aU → x.

17.  Let A be a subset of a topological space X. Then 

x ∈ A if and only if there is a net of points of A converging to x.

Proof  Suppose x ∈ A.  

Let J be the collection of all neighborhoods of x.

Since x ∈ A, U ∩ A is non-empty, for each U ∈ J

i.e. there exists aU ∈ U ∩ A, for each U ∈ J.

By 16., (aU)U ∈ J is a net in A and aU → x.

Conversely, suppose there is a net (aα)α ∈ J of points of A converging to x.

Then for each neighborhood U of x, there exists α ∈ J such that α ⪯ β ⇒ aβ ∈ U.

Hence every neighbourhood U of x contains a point of A and hence x ∈ A. 

18.  Let X and Y be topological spaces, x ∈ X and f : X → Y. Then 

f is continuous at x if and only if for every net (xα) in X converging to x, the net

(f(xα)) converges to f(x).

Proof  Suppose f is continuous at x. And (xα) is a net in X converging to x.

Let V be any neighbourhood of f(x) in Y. Since f is continuous at x, f – 1(V) is open

in X and x ∈ f – 1(V). 

Since xα → x in X, there exists α ∈ J such that α ⪯ β ⇒ xβ ∈ f – 1(V) ⇒ f(xβ) ∈ V.

Hence the net (f(xα)) converges to f(x).

Conversely, suppose for every net (xα) in X converging to x, the net (f(xα)) converges to f(x).

Suppose f is not continuous at x.

Then there exists an open set V in Y such that f(x) ∈ V and f - 1(V) is not open in X.

Let J be the collection of all neighborhoods of x.

Since x ∈ f - 1(V) and f - 1(V) is not open in X, U ⊈ f - 1(V) for all U ∈ J.

(if U ⊆ f - 1(V) for some U then f - 1(V) is open in X)

Hence for each U ∈ J, there exists aU ∈ U such that aU ∉ f - 1(V) 

By 16., (aU)U ∈ J is a net in A and aU → x. 

Since f(aU) ∉ V, for all U ∈ J and V is a neighbourhood of f(x), f(aU) ⇸ f(x).

Which contradicts the hypothesis, 

for all nets (xα) in X converging to x, the net (f(xα)) converges to f(x).

Hence f is continuous at a. 

Oh, Nets are adequate to determine closed sets and continuous functions of

topological spaces.

Not only this but also Nets are adequate to determine Compact Spaces.

To verify this we have to define a subnet of a net. So, we first look into

the definition of a subsequence of a sequence.

19. Recall If (xn) is a sequence in a topological space X and 

n1 < n2 < . . . < nk < . . .

is a strictly increasing sequence of natural numbers then 

(xni) is a subsequence of (xn).

i.e. if f : ℕ → X is a sequence in X and f (n) = xn and 

g : ℕ → ℕ is given by g(i) = ni, i = 1, 2, . . .  such that i ≤ j ⇒ g(i) ≤ g(j) then 

f ◦ g is a subsequence of the sequence f.

And f ◦ g (i) = f(ni) = xni and we say that (xni) is a subsequence of (xn). 

We note that for each i in the codomain ℕ of g, i ≤ g(i) = ni.

i.e. for each i in the codomain ℕ of g, there exists g(i) ∈ g(ℕ) such that i ≤ g(i).

So, g(ℕ) is a subset of ℕ and 

for each i ∈ ℕ, there exists j ∈ g(ℕ) such that i ⪯ j.

In general such a subset of a directed set J is said to be cofinal in J.

20.  If J is a directed set then a subset K of J is said to be cofinal in J 

if for each α ∈ J, there exists β ∈ K such that α ⪯ β.

 

21.  If J is a directed set and K is cofinal in J, then K is non-empty and K is a

directed set(verify).

 

22.  Let f : J → X be a net in X; let f (α) = xα. 

If K is a directed set and g : K → J is a function such that 

(i) i ⪯ j in K ⇒ g(i) ⪯ g(j) in J and (ii) g(K) is cofinal in J, 

then the composite function f ◦ g : K → X is called a subnet of (xα). 

i.e. (xg(i))i ∈ K is a subnet of (xα).

23.  If (xn) is a sequence in a topological space X, then any subsequence (xni) of

(xn) is a subnet of the net (xn). But a subnet of the sequence (xn) need not be a

subsequence.

For,  Since (xni) is a subsequence of (xn), n1 < n2 < . . . < nk < . . .  .

Let K = ℕ and define g : ℕ → ℕ by g(i) = ni, i = 1, 2, . . .  .

Then  i ≤ j ⇒ g(i) ≤ g(j) and since for each i ∈ ℕ, i ≤ g(i) = ni, g(K) is cofinal in ℕ.

And f ◦ g (i) = f(ni) = xni, where f(n) = xn.

Hence (xni) is a subnet of the net (xn).

Suppose K be the ray [1, ∞) of all positive real numbers greater than or equal to 1

and g : K → ℕ be defined by g(x) = [x], the integral part of x. 

Then K is a directed set and g : K → ℕ is a function such that 

(i) x ⪯ y in K ⇒ g(x) ⪯ g(y) in ℕ and (ii) g(K) = ℕ is cofinal in ℕ. 

Hence (xg(i))i ∈ K is a subnet of the sequence (xn).

But (xg(i))i ∈ K is x1, x1, x1, . . . , x2, x2, x2, . . . ,  xn, xn, xn, . . . , . . . . . , 

where each  xi is repeated uncountably. 

Hence the subnet (xg(i))i ∈ K is not a subsequence of the sequence (xn).  

24.  If the net (xα) converges to x, so does any subnet. 

For,  Suppose (xg(i))i ∈ K is a subnet of (xα), where K is a directed set and g : K → J

is a function such that (i) i ⪯ j ⇒ g(i) ⪯ g(j) and (ii) g(K) is cofinal in J.

Let U be a neighborhood of x. 

Since xα → x, there exists α ∈ J such that α ⪯ β ⇒ xβ ∈ U.

Since g(K) is cofinal in J, there exists β ∈ g(K) such that α ⪯ β.  

Clearly, β = g(i) for some i ∈ K.

Now i ⪯ j in K ⇒ β = g(i) ⪯ g(j) in g(K)

                         ⇒ α ⪯ β = g(i) ⪯ g(j) in J

                         ⇒ xg(j) ∈ U.

Hence for each neighborhood U of x,

there exists i ∈ K such that i ⪯ j in K ⇒ xg(j) ∈ U.

i.e. xg(i) → x.

25. Recall : A point x in a topological space X is an accumulation point of the

sequence (xn) in X if each neighborhood U of x contains infinitely many terms

of (xn). 

And if the space X is first countable then x is an accumulation point of the

sequence (xn) if and only if some subsequence of (xn) converges to x.

How to define an accumulation point of a net (xα)α ∈ J? 

Since J is a directed set, we cannot speak of infinitely many terms of (xα).

But suppose a subset U of X contains infinitely many terms of a sequence (xn)

in X. Then for each n ∈ ℕ, there exists m ∈ ℕ such that n ⪯ m and xm ∈ U.

And conversely, suppose for each n ∈ ℕ, there exists m ∈ ℕ such that n ⪯ m

and xm ∈ U. Then U contains infinitely many terms of a sequence (xn).

Hence x is an accumulation point of the sequence (xn) if and only if for each

neighborhood U of x, for each n ∈ ℕ, there exists m ∈ ℕ such that n ⪯ m and

xm ∈ U. 

So, we can define an accumulation point of a net (xα)α ∈ J as:

26.  Let (xα)α ∈ J be a net in X. We say that x is an accumulation point of the net (xα)

if for each neighborhood U of x, for each α ∈ J, there exists β ∈ J such that

α ⪯ β and xβ ∈ U.

i.e. if for each neighborhood U of x, the set of those α for which xα ∈ U is cofinal

in J.

27.  The net (xα) has the point x as an accumulation point if and only if some subnet of

(xα) converges to x. 

Proof  Suppose x is an accumulation point of the net (xα). Then 

for each neighborhood U of x and for each α ∈ J, there exists β ∈ J such that

α ⪯ β and xβ ∈ U.

Let K = {(α, U) / α ∈ J and U is a neighborhood of x containing xα}. 

Define (α, U) ⪯ (β, V) if α ⪯ β and V ⊂ U. 

It can be verified that K is a directed set.

Define g : K → J by g((α, U)) = α.

Then (α, U) ⪯ (β, V) ⇒ g((α, U))  = α ⪯ g((β, V)) = β and V ⊂ U

For each α ∈ J, for each neighborhood U of x there exists β ∈ J such that 

α ⪯ β and xβ ∈ U i.e. (β, U) ∈ K and α ⪯ β = g(β, U) ∈ g(K) .

Hence g(K) is cofinal in J, 

And  (xg((α, U)))(α, U) ∈ K is a subnet of (xα). 

If U is a neighborhood of x and α ∈ J then there exists β ∈ J such that 

α ⪯ β and xβ ∈ U. Hence (β, U) ∈ K. And 

(β, U) ⪯ (𝛾, V) in K ⇒ g((β, U)) = β ⪯ g((𝛾, V)) = 𝛾 and V ⊂ U. 

Since (𝛾, V) ∈ K, xg((𝛾, V)) = x𝛾 ∈ V ⊂ U.

Hence there exists (β, U) ∈ K and (β, U) ⪯ (𝛾, V) in K ⇒ xg((𝛾, V)) = x𝛾 ∈ U.

Therefore the subnet (xg((α, U)))(α, U) ∈ K  of (xα) converges to x.

Conversely, suppose some subnet (xg(i))i ∈ K of (xα) converges to x, 

where K is a directed set and g : K → J is a function such that 

(i) i ⪯ j ⇒ g(i) ⪯ g(j) and (ii) g(K) is cofinal in J.

Let U be a neighborhood of x. Let α ∈ J. 

Since g(K) is cofinal in J, there exists g(k) ∈ g(K) such that α ⪯ g(k).

Since the subnet (xg(i))i ∈ K of (xα) converges to x,

there exists i ∈ K such that i ⪯ j in K ⇒ xg(j) ∈ U.

Since i, k ∈ K and K is a directed set, 

there exists j ∈ K such that i ⪯ j and k ⪯ j in K.

Hence xg(j) ∈ U and g(k) ⪯ g(j) in J.

Since α ⪯ g(k), α ⪯ g(j).

Therefore, for each α ∈ J, there exists β = g(j) ∈ J such that

α ⪯ β and xβ ∈ U.

Hence x is an accumulation point of the net (xα). 

28. X is compact if and only if every net in X has a convergent subnet. 

Proof  Suppose X is compact. 

Let (xα)α ∈ J be a net in X.

For each α ∈ J, let Bα = {xβ | α ⪯ β}.

Since α ⪯ α, xα ∈ Bα and hence Bα is  non-empty.

Since J is a directed set, for α, β ∈ J, 

there exists γ ∈ J such that α ⪯ γ and β ⪯ γ. 

Hence x𝛾 ∈ Bα ∩ Bβ. 

So,{Bα} is a collection of non-empty subsets of X and has the finite intersection

property. Hence {Bα} has the finite intersection property. 

Since X is compact there exists x ∈ ∩ Bα.

Hence for each neighborhood U of x and for each α ∈ J, there exists y ∈ U ∩ Bα.

Since y ∈ Bα, y = xβ, for some β ∈ J such that α ⪯ β and xβ ∈ U.

So, x is an accumulation point of the net (xα). 

Hence by 27., some subnet of (xα) converges to x. 

Therefore, every net in X has a convergent subnet.

Conversely, suppose every net in X has a convergent subnet.

Let 𝓐 be a collection of closed sets having the finite intersection property.

Let 𝓑 be the collection of all finite intersections of elements of 𝓐.

Then 𝓑 is closed under finite intersections and hence by 13.C., 𝓑 is a directed set,

partially ordered by reverse inclusion.

Since each B ∈ 𝓑 is a finite intersections of elements of 𝓐 and 

𝓐 has finite intersection property, each B ∈ 𝓑 is non-empty. 

For each B ∈ 𝓑, fix xB ∈ B.

Then (xB)B ∈ 𝓑 is a net in X. By the hypothesis the net (xB)B ∈ 𝓑 has a convergent

subnet. Suppose the convergent subnet converges to x ∈ X. 

Then by 27., x is an accumulation point of the net (xB)B ∈ 𝓑.  

For each neighbourhood U of x, for each B ∈ 𝓑, there exists C ∈ 𝓑 such that

B ⊇ C and xC ∈ U.

Let A ∈ 𝓐. Then A ∈ 𝓑, hence for each neighbourhood U of x there exists C ∈ 𝓑 

such that A ⊇ C and xC ∈ U.

Since xC ∈ C, xC ∈ A and xC ∈ U. i.e. xC ∈ A ∩ U. 

So, each neighbourhood U of x intersects A.

Hence x ∈ A = A (since each A ∈ 𝓐  is closed).

Thus x ∈ A, for all A ∈ 𝓐.

i.e. x ∈ ∩A ∈ 𝓐 A and ∩A ∈ 𝓐 A is non-empty.

Hence X is compact.

So, 

In topological spaces Nets are adequate to determine not only closed sets and

continuous functions but also to determine Compact Spaces.