Interiors and closures

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 4 March 2014

Interiors and closures

Let X be a topological space and let E⊆X.
The interior of E is the subset Eo of X such that

(a)   Eo is open and Eo⊆E, and
(b)   if U is open and U⊆E then U⊆ Eo.
The closure of E is the subset E‾ of X such that
(a)   E‾ is closed and E‾ ⊇E, and
(b)   if V is closed and V⊇E then V⊇E‾.

Let X be a topological space and let E⊆X.
An interior point of E is a point x∈X such that there exists a neighbourhood N of x such that N⊆E.
A close point of E is a point x∈X such that if N is a neighbourhood of x then N∩E≠∅.

Let X be a topological space. Let E ⊆X.

(a)   The interior of E is the set of interior points of E.
(b)   The closure of E is the set of close points of E.

Proof (of part a).
  1. Let I= {x∈E | xis an interior point of E}.
  2. To show that Eo=I, we show that (aa) I⊆Eo and then that (ab) Eo⊆I.
    1. Let x∈I. Then there exists a neighbourhood N of x with N⊆E.
    2. So there exists an open set U with x∈U⊆N ⊆E.
    3. Since U⊆E and U is open U⊆Eo.
    4. So x⊆Eo.
    5. So I⊆Eo.
    1. We want to show that if x∈Eo then x∈I.
    2. Assume x∈Eo.
    3. Then Eo is open and x∈Eo⊆E.
    4. So x is an interior point of E.
    5. So x∈Eo.
    6. So I⊆Eo.

HW: Let X be a topological space and let E⊆X.

(a) Show that Ec‾=(E∘)c by using the definition of closure.
(b) Show that (Ec)∘=(E‾)c, by taking complements and using (a).
(c) Show that C={x∈X | if N∈𝒩(x) then N∩E≠∅} is the set of close points of E.
(d) Show that Cc=(Ec)∘.
(e) Show that C=E‾.

Proof.

(a)
To show: (aa) (E∘)c is closed and (E∘)c⊇Ec.
(ab) If V is closed and V⊇Ec then V⊇(E∘)c.
(aa) Since E∘ is open, then (E∘)c is closed.
Since E∘⊆E then (E∘)c⊇Ec.
(ab) Assume V is closed and V⊇Ec.
Then Vc is open and Vc⊆E.
So Vc⊆E∘.
So V⊇(E∘)c.
So (E∘)c=Ec‾.
(b)
To show: (Ec)∘=(E‾)c.
To show: ((Ec)∘)c=E‾.
By (a), ((Ec)∘)c= (Ec)c‾= E‾.
(c) By definition of close point C={x∈X | if N∈𝒩(x) then N∩E≠∅} is the set of close points of E.
(d) By definition of C, Cc = { x∈X |  there exists N∈𝒩(x)  such that N∩E=∅ } = { x∈X | there exists  N∈𝒩(x) such that  N⊆Ec } , which is the set of interior points of Ec. Thus Cc=(Ec)∘, by Proposition 1.1(a).
(e)
To show: C=E‾.
To show: Cc=(E‾)c.
By (d) and (b), Cc=(Ec)∘=(E‾)c.

□

Notes and References

These notes follow Bourbaki [Bou, Ch. 1 § 1.6].

The definition of the interior of E is the mathematically precise formulation of "E∘ is the largest open set contained in E". The definition of the closure of E is the mathematically precise formulation of "E‾ is the smallest closed sets containing E". These notes follow Bourbaki [Bou, Ch. I §1 no. 6]. Similar information is treated in [Ru, Ch. 2, 2.18-2.27].

References

[Bou] N. Bourbaki, General Topology, Springer-Verlag, 1989. MR1726779.

[BR] W. Rudin, Principles of mathematical analysis, Third edition, International Series in Pure and Applied Mathematics, McGraw-Hill 1976. MR0385023.

[Ru] W. Rudin, Real and complex analysis, Third edition, McGraw-Hill, 1987. MR0924157.

page history