# Regularity is hereditary

This article gives the statement, and possibly proof, of a topological space property (i.e., regular space) satisfying a topological space metaproperty (i.e., subspace-hereditary property of topological spaces)
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## Statement

Any subset of a regular space is regular under the subspace topology.

## Definitions used

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## Related facts

### Similar facts

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

Other similar facts:

## Facts used

1. T1 is hereditary

## Proof

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

To prove: $A$ is 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 exist disjoint open subsets $U_1, U_2$ of $A$ such that $x \in U_1$ and $C \subseteq U_2$.

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 $A$.
2 $x$ is not in $D$ $x \in A, x \notin C$ Step (1) [SHOW MORE]
3 There exist disjoint open subsets $V_1, V_2$ of $X$ such that $x \in V_1, D \subseteq V_2$. $X$ is regular Steps (1), (2) Step-given combination direct
4 Define $U_1 = V_1 \cap A$ and $U_2 = V_2 \cap A$.
5 $U_1, U_2$ are open subsets of $A$. definition of subspace topology Steps (3), (4) By Step (2), $V_1,V_2$ are open, so by the definition of subspace topology, $U_1, U_2$ are open as per their definitions in Step (3).
6 $U_1, U_2$ are disjoint. Steps (3), (4) follows directly from $V_1,V_2$ being disjoint
7 $x \in U_1, C \subseteq U_2$ $x \in A, C \subseteq A$ Steps (3), (4) By Step (4), $U_1 = V_1 \cap A$. By Step (3), $x \in V_1$, and we are also given that $x\in A$, so $x_1 \in V_1 \cap A = U_1$. Similarly, $C \subseteq V_2$ and $C \subseteq A$, so $C \subseteq V_2 \cap A = U_2$.
8 $U_1, U_2$ are the desired open subsets. Steps (5)-(7) Step-combination, this is exactly what we want to prove.
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## References

### Textbook references

• Topology (2nd edition) by James R. Munkres, More info, Page 196, Theorem 31.2(b), Chapter 4, Section 31