Pasting Lemma

by elementary set theory. Since f is continuous, f –1(C) is closed in A and, therefore, closed in X. Similarly, g –1(C) is closed in B and therefore closed in X. Their union

h –1(C) is thus closed in X.

This Pasting Lemma also holds if A and B are open in X.

When we learn this first we have to think what elementary set theory have to

be used to show h –1(C) = f –1(C) ⋃ g –1(C), and where

“If f(x) = g(x) for every x ∈ A ⋂ B” and “A and B are closed in X” are used.

Yes, since f(x) = g(x) for every x ∈ A ⋂ B, h(x) is uniquely defined for all x ∈ X.

And h –1(C) = h –1(C) ⋂ X = h –1(C) ⋂ (A ⋃ B)

= (h –1(C) ⋂ A) ⋃ (h –1(C) ⋂ B)

(since ⋂ distributes over ⋃, in fact over arbitrary ⋃)

= f –1(C) ⋃ g –1(C) (since h –1(C) ⋂ A = {x ∈ A / h(x) ∈ C}

= {x ∈ A / f(x) ∈ C} = f –1(C) )

Since f –1(C) is closed in A, f –1(C) = A ⋂ D, for some D closed in X.

Since A and D are closed in X, f –1(C) = A ⋂ D is closed in X.

Similarly B is closed in X is used to get g –1(C) is closed in X.

After understanding the proof we think of the case A and B are open in X.

Yes,

If A and B are open in X, then in the proof we have to replace the word “closed”

by “open” in all the places.

After studying this way the natural question have to arise within us is :

Does the lemma hold if X is a union of finite number of closed (or open)

sets and if X is a union of an arbitrary union of closed (or open) sets ?

First we consider the finite union case :

Let X and Y be topological space and X = ⋃i = 1 to n Ai , where Ai , i = 1, 2, . . . , n

are closed in X. Let fi : Ai → Y be continuous, i = 1, 2, . . . , n.

If for all i ≠ j, fi(x) = fj(x) for every x ∈ Ai ⋂ Aj,

then all the fi s combine to give a function h : X → Y, defined by setting

h(x) = fi(x) if x ∈ Ai . Is h continuous ?

And is h continuous ?, if Ai , i = 1, 2, . . . , n are open in X.

Yes, we verify !

Since for all i ≠ j, fi(x) = fj(x) for every x ∈ Ai ⋂ Aj,

h(x) is uniquely defined for all x ∈ X.

Let C be a closed subset of Y. Now

h –1(C) = h –1(C) ⋂ X = h –1(C) ⋂ (⋃i = 1 to n Ai)

= ⋃i = 1 to n(h –1(C) ⋂ Ai)

(since ⋂ distributes over ⋃, in fact over arbitrary ⋃)

= ⋃i = 1 to n f i–1(C) (since h –1(C) ⋂ Ai = {x ∈ Ai / h(x) ∈ C}

= {x ∈ Ai / fi(x) ∈ C} = fi –1(C) )

For each i, since fi is continuous, fi –1(C) is closed in Ai .

Hence fi –1(C) = Ai ⋂ Di, for some Di closed in X.

Since Ai and Di are closed in X, fi –1(C) = Ai ⋂ Di is closed in X.

Therefore h –1(C) = ⋃i = 1 to n f i–1(C) is closed in X,since a finite union of closed sets in X is closed in X.

If Ai, i = 1, 2, . . . , n are open in X, then all the above are trueif we replace the word “closed” by “open” in all the places.

So, Pasting lemma holds in the finite union case :

Let X and Y be topological space and X = ⋃i = 1 to n Ai, where Ai , i = 1, 2, . . . , n

are closed in X. Let fi : Ai → Y be continuous, i = 1, 2, . . . , n.

If for all i ≠j, fi(x) = fj(x) for every x ∈ Ai ⋂ Aj,

then all the fi s combine to give a continuous function h : X → Y,

defined by setting h(x) = fi(x) if x ∈ Ai .

And this Pasting Lemma also holds if Ai , i = 1, 2, . . . , n are open in X.

Now we consider the arbitrary union case :

Let X and Y be topological space and X = ⋃𝞪 ∈ I A𝞪 where A𝞪 , 𝞪 ∈ I

are closed in X. Let f𝞪 : A𝞪 → Y be continuous, 𝞪 ∈ I.

If for all 𝞪, 𝜷 ∈ I, f𝞪(x) = f𝜷(x) for every x ∈ A𝞪 ⋂ A𝜷,

then all the f𝞪 s combine to give a function h : X → Y, defined by setting

h(x) = f𝞪(x) if x ∈ A𝞪 . Is h continuous ?

And is h continuous, if A𝞪 , 𝞪 ∈ I are open in X ?.

Yes, we verify!

Since for all 𝞪, 𝜷 ∈ I, f𝞪(x) = f𝜷(x) for every x ∈ A𝞪 ⋂ A𝜷,h(x) is uniquely defined for all x ∈ X.

Let C be a closed subset of Y. Now

h –1(C) = h –1(C) ⋂ X = h –1(C) ⋂ (⋃𝞪 ∈ I A𝞪)

= ⋃𝞪 ∈ I(h –1(C) ⋂ A𝞪)

(since ⋂ distributes over arbitrary ⋃)

= ⋃𝞪 ∈ I f 𝞪–1(C)

(since h –1(C) ⋂ A𝞪 = {x ∈ A𝞪 / h(x) ∈ C}

= {x ∈ A𝞪 / f𝞪(x) ∈ C} = f𝞪 –1(C) )

For each 𝞪, since f𝞪 is continuous, f𝞪 –1(C) is closed in A𝞪 .

Hence f𝞪 –1(C) = A𝞪 ⋂ D𝞪, for some D𝞪 closed in X.

Since A𝞪 and D𝞪 are closed in X, f𝞪 –1(C) = A𝞪 ⋂ D𝞪 is closed in X.

But since arbitrary union of closed sets need not be closed in X,

h –1(C) = ⋃𝞪 ∈ I f 𝞪–1(C) need not be closed in X.

So, the Pasting Lemma need not hold for arbitrary union of closed sets.

However, if A𝞪, 𝞪 ∈ I are open in X, then all the above steps holds

if we replace the word “closed” by “open” in all the places.

And since arbitrary union of open sets is open in X,

h –1(C) = ⋃𝞪 ∈ I f 𝞪–1(C) is open in X.

So, Pasting Lemma holds for arbitrary union of open sets :

Let X and Y be topological space and X = ⋃𝞪 ∈ I A𝞪 where A𝞪 , 𝞪 ∈ I

are open in X. Let f𝞪 : A𝞪 → Y be continuous, 𝞪 ∈ I.

If for all 𝞪, 𝜷 ∈ I, f𝞪(x) = f𝜷(x) for every x ∈ A𝞪 ⋂ A𝜷,

then all the fi s combine to give a continuous function h : X → Y, defined by setting

h(x) = f𝞪(x) if x ∈ A𝞪 .

After this observation Think what are the possibilities to conclude.

First we have to get an example for Pasting Lemma does not hold for arbitrary

union of closed sets and second we can think

for additional conditions required so that h –1(C) = ⋃𝞪 ∈ I f 𝞪–1(C) is closed in X.

Let X = {1/n / n = 1, 2, . . . } ⋃ {0}.

Then X = ⋃n {1/n} ⋃ {0}, {1/n} and {0} are closed in X.

Let fn : {1/n} → 𝓡 be defined by fn(1/n) = n and f0 : {0} → 𝓡 be defined by

f0(0) = 0. Then fn and f0 are continuous.

But f : X → 𝓡 defined by f(1/n) = n and f(0) = 0 is not continuous,

since 1/n → 0 in X but f (1/n) = n not converge to 0.

We note that we are not able to show that h –1(C) = ⋃𝞪 ∈ I f 𝞪–1(C) is closed in X,

since arbitrary union of closed sets need not be closed in X.

So, we show that h –1(C) is closed in X in some other way,

of course with additional conditions.

Yes, we can show X – h –1(C) is open in X.

i.e. for each x ∉ h –1(C), we have to get a neighbourhood U of x in X such that

U ⋂ h –1(C) = ∅.

Since we have to find additional conditions on A𝞪 s or X or Y,

we write h –1(C) in terms of A𝞪.

Let B = h –1(C).

Then B ⋂ A𝞪 = f 𝞪–1(C) is closed in A𝞪 for each 𝞪 ∈ I and B = ⋃𝞪 ∈ I (B ⋂ A𝞪).

We have to show that X – B is open.

Let x ∈ X – B.

We have to find conditions so that there is a open set containing x and

contained in X.

How to find it ? We start with an open set containing x.

Suppose U is open in X and x ∈ U. Then

U ⋂ B = U ⋂ (⋃𝞪 ∈ I (B ⋂ A𝞪)) = ⋃𝞪 ∈ I (U ⋂ B ⋂ A𝞪)

Since x ∈ X = ⋃𝞪 ∈ I A𝞪 , U ⋂ A𝞪 ≠ ∅ for at least one 𝞪 ∈ I.

Suppose for such 𝞪, U ⋂ B ⋂ A𝞪 ≠ ∅ then U ⋂ B ≠ ∅.

How to deal 𝞪 s with U ⋂ B ⋂ A𝞪 ≠ ∅ to get another neighbourhood V𝞪 of x

so that V𝞪 ⋂ B ⋂ A𝞪 = ∅.

Yes, V𝞪 should be open containing x and contained in X – (B ⋂ A𝞪).

Since B ⋂ A𝞪 = f 𝞪–1(C) is closed in X, X – (B ⋂ A𝞪) is open in X. we use it.

If we take V𝞪 = U ⋂ (X – (B ⋂ A𝞪)) then V𝞪 = U ⋂ (X – (B ⋂ A𝞪)) is open in X.

And since x ∉ B, x ∉ B ⋂ A𝞪 hence x ∈ X – (B ⋂ A𝞪) and hence x ∈ V𝞪.

So V𝞪 is a neighbourhood of x and V𝞪 ⋂ (B ⋂ A𝞪) = ∅.

Suppose J = {𝞪 ∈ I / U ⋂ (B ⋂ A𝞪) ≠ ∅}.

And V = U ⋂ ⋂𝞪 ∈ J (X – (B ⋂ A𝞪)) = ⋂𝞪 ∈ J (U ⋂ (X – (B ⋂ A𝞪))) = ⋂𝞪 ∈ J V𝞪.

Then x ∈ V and V ⋂ (B ⋂ A𝞪) = ∅, for all 𝞪 ∈ J.

Since for all 𝞪 ∈ I – J, U ⋂ (B ⋂ A𝞪) = ∅, V ⋂ (B ⋂ A𝞪) = ∅, for all 𝞪 ∈ I – J.

Hence V ⋂ (B ⋂ A𝞪) = ∅, for all 𝞪 ∈ I.

And hence V ⋂ B = V ⋂ (⋃𝞪 ∈ I (B ⋂ A𝞪)) = ⋃𝞪 ∈ I (V ⋂ B ⋂ A𝞪) = ∅.

Since only a finite intersection of open set is open,

V is open in X only when J is a finite set. So,

If J is a finite set then V is a neighbourhood of x and V ⋂ B = ∅.

Hence, if for each x ∉ B there exists a neighbourhood U of x such that

U ⋂ (B ⋂ A𝞪) ≠ ∅ only for a finitely many 𝞪 ∈ I then there is a neighbourhood

V of x such that V ⋂ B = ∅.

i.e. if for each x ∉ B there exists a neighbourhood U of x such that

U ⋂ (B ⋂ A𝞪) ≠ ∅ only for a finitely many 𝞪 ∈ I then X – B is open in X and

hence B = h –1(C) is closed in X.

But h is continuous if it is true for all closed sets C in Y.

Hence if for all closed sets C in Y, for each x ∉ B = h –1(C) there exists

a neighbourhood U of x such that U ⋂ (B ⋂ A𝞪) ≠ ∅ only for a finitely many

𝞪 ∈ I then h is continuous.

Think how to state this necessary condition on all closed sets C in Y

by conditions on X and Y.

First we note that the part of the condition “for all closed sets C in Y, for each

x ∉ B = h –1(C) there exists a neighbourhood U of x” follows from the

condition “if for each x ∈ X there exists a neighbourhood U of x”.

Think how to fulfil the later part of the condition “ such that

U ⋂ (B ⋂ A𝞪) ≠ ∅ only for a finitely many 𝞪 ∈ I ” without using B.

Yes, if U ⋂ A𝞪 ≠ ∅ only for a finitely many 𝞪 ∈ I then for any B ⊆ X,

U ⋂ (B ⋂ A𝞪) ≠ ∅ only for a finitely many 𝞪 ∈ I.

Hence if for each x ∈ X there exists a neighbourhood U of x such that

U ⋂ A𝞪 ≠ ∅ only for a finitely many 𝞪 ∈ I then h is continuous.

Oh! we have succeeded !!.

And we can think of calling “the family of sets { A𝞪 / 𝞪 ∈ I} in X

such that for each x ∈ X there exists a neighbourhood U of x such that

U ⋂ A𝞪 ≠ ∅ only for a finitely many 𝞪 ∈ I” by some special name.

But in Topology by Munkres such family is called locally finite and

in Topology by James Dugundji it is called neighbourhood finite or nbd-finite.

What does it mean ?

Of course we have taken good old concepts.

Already, very early, the Pasting lemma has been learned in this direction.

Anyhow, We can feel proud.

For the sake of completion, we give the definition of locally finite family of sets

and state and prove the Pasting Lemma for arbitrary union of closed sets.

Let X be a topological space. A family { A𝞪 / 𝞪 ∈ I} of subsets in X is said to be

locally finite or neighbourhood finite or nbd-finite if for each x ∈ X there exists

a neighbourhood U of x such that U ⋂ A𝞪 ≠ ∅ only for a finitely many 𝞪 ∈ I.

Pasting Lemma for arbitrary union of closed sets :

Let X and Y be topological space and X = ⋃𝞪 ∈ I A𝞪 where A𝞪 , 𝞪 ∈ I

are closed in X. Let f𝞪 : A𝞪 → Y be continuous, 𝞪 ∈ I and

for all 𝞪, 𝜷 ∈ I, f𝞪(x) = f𝜷(x) for every x ∈ A𝞪 ⋂ A𝜷,

Let h : X → Y be defined by h(x) = f𝞪(x) if x ∈ A𝞪 .

If { A𝞪 / 𝞪 ∈ I} is locally finite or neighbourhood finite then h is continuous.

Proof

Since for all 𝞪, 𝜷 ∈ I, f𝞪(x) = f𝜷(x) for every x ∈ A𝞪 ⋂ A𝜷,

h(x) is uniquely defined for all x ∈ X.

Let C be a closed subset of Y. Now

h –1(C) = h –1(C) ⋂ X = h –1(C) ⋂ (⋃𝞪 ∈ I A𝞪)

= ⋃𝞪 ∈ I(h –1(C) ⋂ A𝞪)

(since ⋂ distributes over arbitrary ⋃)

= ⋃𝞪 ∈ I f 𝞪–1(C)

(since h –1(C) ⋂ A𝞪 = {x ∈ A𝞪 / h(x) ∈ C}

= {x ∈ A𝞪 / f𝞪(x) ∈ C} = f𝞪 –1(C) )

For each 𝞪, since f𝞪 is continuous, f𝞪 –1(C) is closed in A𝞪 .

Hence f𝞪 –1(C) = A𝞪 ⋂ D𝞪, for some D𝞪 closed in X.

Since A𝞪 and D𝞪 are closed in X, f𝞪 –1(C) = A𝞪 ⋂ D𝞪 is closed in X.

Let B = h –1(C).

Then B = ⋃𝞪 ∈ I (B ⋂ A𝞪), where B ⋂ A𝞪 = f 𝞪–1(C) is closed in A𝞪 for each 𝞪 ∈ I.

To prove B is closed we show that X – B is open.

Let x ∈ X - B.

Since { A𝞪 / 𝞪 ∈ I} is locally finite, there exists a neighbourhood U of x such that

U ⋂ A𝞪 ≠ ∅ only for a finitely many 𝞪 ∈ I.

Since B ⋂ A𝞪 ⊆ A𝞪, U ⋂ (B ⋂ A𝞪) ≠ ∅ only for a finitely many 𝞪 ∈ I.

Suppose J = {𝞪 ∈ I / U ⋂ (B ⋂ A𝞪) ≠ ∅}.

Since B ⋂ A𝞪 = f 𝞪–1(C) is closed in X, X – (B ⋂ A𝞪) is open in X and hence

V𝞪 = U ⋂ (X – (B ⋂ A𝞪)) is open in X.

And since x ∉ B, x ∉ B ⋂ A𝞪 hence x ∈ X – (B ⋂ A𝞪) and hence x ∈ V𝞪.

So V𝞪 is a neighbourhood of x and V𝞪 ⋂ (B ⋂ A𝞪) = ∅.

Let V = U ⋂ ⋂𝞪 ∈ J (X – (B ⋂ A𝞪)) = ⋂𝞪 ∈ J (U ⋂ (X – (B ⋂ A𝞪))) = ⋂𝞪 ∈ J V𝞪.

Then since J is a finite set, V is open in X, x ∈ V.

Since V𝞪 ⋂ (B ⋂ A𝞪) = ∅ and V ⊆ V𝞪, V ⋂ (B ⋂ A𝞪) = ∅, for all 𝞪 ∈ J.

Since for all 𝞪 ∈ I – J, U ⋂ (B ⋂ A𝞪) = ∅ and V ⊆ U,

V ⋂ (B ⋂ A𝞪) = ∅, for all 𝞪 ∈ I – J.

Hence V ⋂ (B ⋂ A𝞪) = ∅, for all 𝞪 ∈ I.

And hence V ⋂ B = V ⋂ (⋃𝞪 ∈ I (B ⋂ A𝞪)) = ⋃𝞪 ∈ I (V ⋂ (B ⋂ A𝞪)) = ∅.

Hence X – B is open in X.

i.e. B = h –1(C) is closed in X.

Hence h is continuous.

We note that in the above proof we have proved the following :

1. If {A𝞪 / 𝞪 ∈ I} is locally finite family of closed subsets

of a topological space X then ⋃𝞪 ∈ I A𝞪 is closed in X.

Oh ! arbitrary union of closed sets A𝞪 is closed if {A𝞪 / 𝞪 ∈ I} is locally finite.

2. If {A𝞪 / 𝞪 ∈ I} is locally finite family of subsets of a topological space X then

for any subset B of X, the family {B ⋂ A𝞪 / 𝞪 ∈ I} is also locally finite.

3. Let {A𝞪 / 𝞪 ∈ I} be locally finite family of subsets of a topological space X and

X = ⋃𝞪 ∈ I A𝞪 . And B be a subset of X.

If B ⋂ A𝞪 is closed in X for each 𝞪 ∈ I then B is closed in X.

Understanding what we learn and raising questions lead to more results.

The questions we raised and Mathematical Thinking lead to the

generalisation of Pasting Lemma.

This is a way of learning the Pasting Lemma.

See how enjoyable Mathematics learning is.

Learners are advised to use this post to develop their mathematical thinking

in this or in any other ways.

With this experience the learner can enjoy the beauties of Pure Mathematics.

Learn and Enjoy Mathematics.

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