General Topology — Part 4 (Urysohn’s Lemma)

8 min readJan 21, 2022

What is Urysohn’s Lemma ?

From previous article, we have learnt many essential topological concepts. Now, you have a fair amount of topology under your belt, you are ready to tackle a deep theorem in topology:

Urysohn’s lemma
A topological space (X, τ) is a Normal Space iff for every two disjoint nonempty closed subsets A, B ⊆ X there is a continuous function f: X → [0,1] s.t. f(A)=0 and f(B)=1

The Normal Space is defined as the topological space (X, τ) that for every two disjoint closed subsets A, B ⊆ X having disjoint open neighbourhoods:
∀A,B⊆ X, Ac, Bc ∈ τ, A⋂B=ø ⇒∃o1, o2 ∈ τ s.t. A ⊆ o1 and B ⊆ o2 and o1⋂o2=ø

This lemma is named after the Russian mathematician Pavel Samuilovich Urysohn. This lemma is famous and widely used since it can be used to construct continuous function. One application example is in the proof of Riesz representation theorem, which is another famous theorem in functional analysis.

Long proof of Urysohn’s Lemma

Yes, the proof is pretty long and it is a test of your patience. But, it is beautiful and I assure that you will get aha moment after reading Urysohn’s proof. Let’s start our journey from the ‘if’ part of this lemma.

‘If’ part-
∀A,B ⊆ X, Ac,Bc ∈ τ, A⋂B=ø
⇒ there is a continuous function f: X→[0,1] s.t. f(A)=0 and f(B)=1

[0,1] is a subspace of R [ See Part 1 ]
⇒ [0,0.5) and (0.5,1] are open sets of [0,1] [ See Part 1 ]
⇒ o1=f-1([0,0.5)), o2=f-1((0.5,1]) where o1,o2∈τ [ By f is continuous, see Part 2 ] and
A ⊂ o1 and B ⊂ o2 [ By f-1(0)=A, f-1(1)=B ]

[0,0.5)⋂(0.5,1]=ø
⇒ o1⋂o2 = f-1([0,0.5))⋂f-1((0.5,1]) = f-1([0,0.5)⋂(0.5,1]) = f-1(ø) = ø [ See Part 1 Set Theory Review ]
⇒ (X,τ) is a Normal Space

For the ‘Only if’ part, we need to prove it by constructing a continuous function. OK, then how? This is what attracts me to study this lemma since it seems impossible to construct a function without given any basis function! What we know is that the topological space τ is a Normal one, f(A)=0 and f(B)=1 for every two disjoint closed subsets A, B.

Construction of Urysohn Function

Urysohn cleverly defines the function as follows:
f(x) = inf{p ∈ D⋂[0,1]| x ∈ o[p]} when x ∈ X\B
= 1 otherwise

where D is the dyadic rational number set, o[ ] is a family of nesting open sets in topological space (X, τ) indexed by a dyadic rational number and inf denotes the infimum. Before I explain how these nesting open sets can be created, I explain the following lemma first:

Lemma A
For any closed set A ⊆ X and an open set C ∈ τ where A ⊆ C, there is an open set o∈τ s.t. A ⊆ o ⊆ Cl(o) ⊆ C

Proof-
C ∈ τ
⇒ Cc is closed
⇒ A⋂Cc = A\C = ø [ By A⊂C, see Part 1]
⇒ ∃o1, o2 ∈ τ s.t. A ⊆ o1 and Cc ⊆ o2 and o1⋂o2=ø [ By definition of Normal Space ]
⇒ ∃o1, o2 ∈ τ s.t. A ⊆ o1 and o2c ⊆ C and o1 ⊆ o2c [ See Part 1 Set Theory Review ]
o2c is closed and o1 ⊆ o2c
⇒ Cl(o1) ⊆ o2c [ By Cl2, see Part 3 ]
⇒ o1 ⊆ Cl(o1) ⊆ C [ By o2c ⊂ C and Cl4, see Part 3 ]
⇒ ∃o1 ∈ τ s.t. A ⊆ o1 ⊆ Cl(o1) ⊆ C

This means that in Normal Space, for any open set B containing a closed set A, we can find another smaller open set o to contain A and do this recursively. Just like the Russian dolls, you can find a smaller one contained inside. In this way, a family of nesting open sets can be created. Then, we index these open sets by Dyadic rational number.

What is dyadic rational number ?

Dyadic rational number is a number in the form of m/2^n where m is an integer and n is a positive integer. Define a subset of this number within [0,1] as follows:
D(n) = {m/2^n | m, n ∈ N and 0 ≤m≤2^n-1 and n≥0}

We index the open set by D(n) and the following property is found hold:
∀p, q∈D(n) p<q ⇒ Cl(o[p]) ⊆ o[q] for n=0,1,2… — — — (Δ)

Now, we index open set X\B by 1:
A ⊆ o[1]
Apply the Lemma A, we get:
A ⊆ o[0] ⊆ Cl(o[0]) ⊆ o[1]
Apply the Lemma A again, we get:
A ⊆ o[0] ⊆ Cl(o[0]) ⊆ o[1/2] ⊆ Cl(o[1/2]) ⊆ o[1]
A ⊆ o[0] ⊆ Cl(o[0]) ⊆ o[1/4] ⊆ Cl(o[1/4]) ⊆ o[1/2] ⊆ Cl(o[1/2]) ⊆ o[3/4] ⊆ Cl(o[3/4]) ⊆ o[1]
and so on….

Let us prove this property Δ is true by induction.
Proof-
When n = 0;
A ⊆ o[0] ⊆ Cl(o[0]) ⊆ o[1]
⇒ So the statement is true for n = 0

Suppose the statement is true for n = k:
∀p,q∈D(k) p<q⇒ Cl(o[p]) ⊆ o[q]

When n = k+1:
Cl(o[m/2^k]) ⊆ o ⊆ Cl(o) ⊆ o[(m+1)/2^k] [By lemma A ]
Index o by (2m+1)/2^(k+1)∈D(k+1):
⇒ Cl(o[m/2^k]) ⊆ o[(2m+1)/2^(k+1)]
⊆ Cl((2m+1)/2^(k+1)) ⊆ o[(m+1)/2^k]
⇒ Cl(o[2m/2^(k+1)]) ⊆ o[(2m+1)/2^(k+1)]
⊆ Cl((2m+1)/2^(k+1)) ⊆ o[(2m+2)/2^(k+1)]
⇒ Statement is true when p and q are consecutive numbers in D(k+1)
When p=2m/2^(k+1) and q=2n/2^(k+1)
⇒ p,q ∈ D(k)
⇒ Statement is also true when p and q are in form 2m/2^(k+1)
⇒ Statement is true when p and q ∈ D(k+1)
Hence, by induction, the statement is true for n=0,1,..

Now, by using Lemma A and let n goes to infinity, we can construct a family of nesting open sets:

A ⊆ o[0] ⊆ Cl(o[0]) ⊆ .. ⊆ o[1/4] ⊆ Cl(o[1/4]) ⊆ .. ⊆ o[1/2] ⊆ Cl(o[1/2]) ⊆ ..⊆ o[3/4] ⊆ Cl(o[3/4]) ⊆ ..⊆ o[1]

D(n) approaches to D⋂[0,1] when n approaches to ∞. If n does not approach to ∞, the range of Urysohn Function will be discrete numbers. The set D⋂[0,1] actually is a Dense Set in [0, 1], that is, for any real number in [0, 1], there exists a number in D⋂[0,1] which is close to this real number. Please refer to my previous article to recap the definition of a Dense Set.

General Topology — Part 3

In this article, I will discuss some essential topological concepts that are closely related: Neighbourhood, Limit…

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Proof-
∀x ∈ [0,1]
⇒ x ∈ [(m-1)/2^n, m/2^n] for some m∈N

∀ε>0
Choose n∈N s.t. ε > 1/2^n or n > -log2(ε)
x + ε > (m-1)/2^n + 1/2^n = m/2^n
⇒ x + ε > m/2^n >= x

x — ε < m/2^n — 1/2^n = (m-1)/2^n
⇒ x — ε < (m-1)/2^n <= x
⇒ (x-ε, x+ε)⋂D ≠ ø
⇒ D⋂[0,1] is dense in [0,1] [ By DS3 ]

‘Only If’ part-
For any x∈X, the set {p ∈ D| x ∈ o[p]} is bounded below by 0 and the set is not empty, infimum exists by Greatest Lower Bound Axiom. (Actually, this can be proved by using Dedekind cut) Also, the infimum is unique and f(x) is defined for all x∈X, f(x) is thus a well-defined function.

OK. We are now at the final stage. What we need to prove is f(x) satisfies the following 2 conditions:

1. f(A)=0 and f(B)=1
2. f(x) is continuous

Proof-
1. f(A)=0 and f(B)=1:
∀x∈A, x∈o(0) [ By A ⊂ o[0]]
⇒ f(x) = inf{p∈D| x∈o(p)} = 0
∀x∈B, f(x) = 1 [ By defintion of f ]

2. f is continuous:

∀x ∈ Cl(o[p]) ⇒ x ∈ o[q] for all q > p ⇒ f(x) ≤ p — — — (*)
∀x ∉ o[p] ⇒ x ∉ o[q] for all q < p ⇒ f(x) ≥ p — — — (**)

For subspace [0, 1] of R, its sub-bases is [0, b) and (a, 1] where 0<a<b<1. Please refer to my previous article to recap the definition of sub-basis.

General Topology — Part 1

What is General Topology ?

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∀x∈f-1([0, b))
⇒ 0 ≤ f(x) < b
⇒ ∃p∈D s.t. f(x)<p<b [ D is dense in [0,1] ]
⇒ x∈o[p] where p<b [ By contrapositive of **]
⇒ x∈⋃{o[p]| p∈D and p<b}

∀x∈⋃{o[p]| p∈D and p<b}
⇒ ∃p∈D s.t. x∈o[p] and p < b
⇒ inf{p∈D| x∈o[p]} ≤ p < b [ Since inf is a lower bound of p ]
⇒ 0 ≤ f(x) < b [ Since f(x) is bounded below by 0 ]
⇒ x∈f-1([0, b))
⇒ f-1([0, b)) = ⋃{o[p]| p∈D and p<b} ∈ τ [ By topology axiom T2 ]

∀x∈f-1((a, 1])
⇒ a < f(x) ≤ 1
⇒ ∃p∈D s.t. a < p < f(x) [ D is dense in [0,1] ]
⇒ x∈X\Cl(o[p]) where p > a [ By contrapositive of * ]
⇒ x∈⋃{X\Cl(o[p])| p∈D and p>a}

∀x∈⋃{X\Cl(o[p])| p∈D and p>a}
⇒ ∃p∈D s.t. x∉Cl(o[p]) and p > a
⇒ ∃q∈D s.t. a < p < q and Cl(o[p]) ⊆ o[q] and x∈o[q] [ By Δ ]
⇒ a < inf{q∈D| x∈o(q)} [ Since inf is the greatest lower bound ]
⇒ a < f(x) ≤ 1 [ Since f(x) is bounded above by 1 ]
⇒ x∈f-1((a, 1])
⇒ f-1((a, 1]) = ⋃{X\Cl(o(p))| p∈D and p>a} ∈ τ [ By topology axiom T2 ]

Since f-1([0, b)) and f-1((a, 1]) are open set, f is continuous. Please refer to my previous article to recap the definition of continuous function.

General Topology — Part 2

ε-δ Definition of Continuous Function

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The Urysohn’s lemma is finally proved !!

Conclusion

The Urysohn’s lemma is discussed in this article. The proof of this lemma is quite long but we have seen how the topological concepts that we learnt in previous articles can be applied altogether to prove this lemma. I hope that you enjoy this journey even if it is quite brain intensive.😄