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 EX.
The interior of E is the subset Eo of X such that

(a)   Eo is open and EoE, and
(b)   if U is open and UE 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 VE then VE.

Let X be a topological space and let EX.
An interior point of E is a point xX such that there exists a neighbourhood N of x such that NE.
A close point of E is a point xX such that if N is a neighbourhood of x then NE.

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= {xE | xis an interior point of E}.
  2. To show that Eo=I, we show that (aa) IEo and then that (ab) EoI.
    1. Let xI. Then there exists a neighbourhood N of x with NE.
    2. So there exists an open set U with xUN E.
    3. Since UE and U is open UEo.
    4. So xEo.
    5. So IEo.
    1. We want to show that if xEo then xI.
    2. Assume xEo.
    3. Then Eo is open and xEoE.
    4. So x is an interior point of E.
    5. So xEo.
    6. So IEo.

HW: Let X be a topological space and let EX.

(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={xX|ifN𝒩(x)thenNE} 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)cEc.
(ab) If V is closed and VEc then V(E)c.
(aa) Since E is open, then (E)c is closed.
Since EE then (E)cEc.
(ab) Assume V is closed and VEc.
Then Vc is open and VcE.
So VcE.
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={xX|ifN𝒩(x)thenNE} is the set of close points of E.
(d) By definition of C, Cc = { xX| there existsN𝒩(x) such thatNE= } = { xX|there exists N𝒩(x)such that NEc } , 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.

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