Complete regularity is hereditary

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This article gives the statement, and possibly proof, of a topological space property (i.e., completely regular space) satisfying a topological space metaproperty (i.e., subspace-hereditary property of topological spaces)
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This article gives the statement, and possibly proof, of a basic fact in topology.

Statement

Any subset of a completely regular space is completely regular in the subspace topology.

Definitions used

Term Definition used
completely regular space A space X is completely regular if it is a T1 space and given any point x \in X and closed subset C \subseteq X such that x \notin C, there exists a continuous mapf:X \to [0,1] such that f(x) = 0 and f(a) = 1 for all a \in C.
subspace topology For a subset A of the space X, the subspace topology on A is defined as follows: a subset of A is open in A iff it can be expressed as the intersection with A of an open subset of X.
Also, a subset of A is closed in A iff it can be expressed as the intersection with A of a closed subset of X.

Related facts

Similar facts

Property Proof that is is subspace-hereditary
Hausdorff space Hausdorffness is hereditary
regular space regularity is hereditary
Urysohn space Urysohn is hereditary
T1 space T1 is hereditary

Other similar facts:

Opposite facts

Facts used

  1. T1 is hereditary

Proof

Given: A topological space X, a subset A of X. X is completely regular.

To prove: A is completely regular.

Proof: The T1 property for A follows from Fact (1). It thus suffices to show the separation property for A, i.e., that any point and disjoint closed subset in A can be separated by a continuous function. In other words, we want to prove the following.

To prove (specific): For any point x \in A and any closed subset C of A such that x \notin A, there exists a continuous function from A to [0,1] such that f(x) = 0 and f(a) = 1 for all a \in C.

Proof (specific):

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 There exists a closed subset D of X such that C= D \cap A definition of subspace topology C is closed in A. Direct from definition. Note: We can, for concreteness, take D = \overline{C}, the closure of C in X -- this is the smallest D that works.
2 x is not in D x \in A, x \notin C Step (1) [SHOW MORE]
3 There exists a continuous function g:X \to [0,1] such that g(x) = 0 and g(a) = 1 for all a \in D. X is completely regular Steps (1), (2) Step-given combination direct
4 Define f as the restriction of g to A. In other words, if i: A \to X is the inclusion mapping, then f = g \circ i. Then, f is a continuous function from A to [0,1]. definition of subspace topology (the inclusion mapping from a subset with the subspace topology is continuous), a composite of continuous maps is continuous Step (3) Step-given combination direct
5 The function f defined in Step (4) has the property that f(x) = 0 and f(a) = 1 for all a \in C. Steps (1), (3), (4) The condition on f follows directly from the condition on g in Step (3) and the fact that C = D \cap A, so C \subseteq D.
This proof uses a tabular format for presentation. Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

References

Textbook references

  • Topology (2nd edition) by James R. Munkres, More info, Page 211-212, Theorem 33.2, Chapter 4, Section 33