Hausdorffness is hereditary

This article gives the statement, and possibly proof, of a topological space property (i.e., Hausdorff space) satisfying a topological space metaproperty (i.e., subspace-hereditary property of topological spaces)
View all topological space metaproperty satisfactions | View all topological space metaproperty dissatisfactions
Get more facts about Hausdorff space |Get facts that use property satisfaction of Hausdorff space | Get facts that use property satisfaction of Hausdorff space|Get more facts about subspace-hereditary property of topological spaces

Statement

Property-theoretic statement

The property of topological spaces of being Hausdorff, is hereditary.

Verbal statement

Any subspace of a Hausdorff space is Hausdorff, in the subspace topology.

Definitions used

Hausdorff space

Further information: Hausdorff space

A topological space $X$ is Hausdorff if given distinct points $a,b\in X$ there exist disjoint open subsets $U,V$ containing $a,b$ respectively.

Subspace topology

Further information: subspace topology

If $A$ is a subset of $X$, we declare a subset $V$ of $A$ to be open in $A$ if $V=U\cap A$ for an open subset $U$ of $X$.

Proof

Given: A topological space $X$, a subset $A$ of $X$. Two distinct points $x_1,x_2\in A$.

To prove: There exist disjoint open subsets $U_1,U_2$ of $A$ such that $x_1\in U_1,x_2\in U_2$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$x_1,x_2$ are distinct points of $X$		$x_1,x_2$ are distinct points of $A$ $A\subseteq X$
2	There exist disjoint open subsets $V_1,V_2$ of $X$ such that $x_1\in V_1,x_2\in V_2$.		$X$ is Hausdorff	Step (1)	Step-given direct
3	Define $U_1=V_1\cap A$ and $U_2=V_2\cap A$.
4	$U_1,U_2$ are open subsets of $A$.	definition of subspace topology		Steps (2), (3)	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).
5	$U_1,U_2$ are disjoint.			Steps (2), (3)	follows directly from $V_1,V_2$ being disjoint
6	$x_1\in U_1,x_2\in U_2$		$x_1,x_2\in A$	Steps (2), (3)	By Step (3), $U_1=V_1\cap A$. By Step (2), $x_1\in V_1$, and we are also given that $x_1\in A$, so $x_1\in V_1\cap A=U_1$. Similarly, $x_2\in U_2$.
7	$U_1,U_2$ are the desired open subsets of $A$.			Steps (4)-(6)	Step-combination direct, it's what we want to prove.

References

Textbook references

• Topology (2nd edition) by James R. Munkres, ^{More info}, Page 100, Theorem 17.11, Page 101, Exercise 12 and Page 196 (Theorem 31.2 (a))