Hausdorffness is hereditary: Difference between revisions

Latest revision as of 20:37, 30 May 2016

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)
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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.

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 100, Theorem 17.11, Page 101, Exercise 12 and Page 196 (Theorem 31.2 (a))

@@ Line 1: / Line 1: @@
-{{topospace metaproperty satisfaction}}
+{{topospace metaproperty satisfaction|
+property = Hausdorff space|
-{{basic fact}}
+metaproperty = subspace-hereditary property of topological spaces}}
+[[Difficulty level::1| ]]
 ==Statement==
-===Property-theoretic statement===
+Any subspace of a [[Hausdorff space]] is Hausdorff in the [[subspace topology]].
-The [[property of topological spaces]] of being [[Hausdorff space|Hausdorff]], is [[hereditary property of topological spaces|hereditary]].
-===Verbal statement===
-Any subspace of a [[Hausdorff space]] is Hausdorff, in the [[subspace topology]].
 ==Definitions used==
@@ Line 29: / Line 23: @@
 ==Proof==
-===Proof outline===
+'''Given''': A topological space <math>X</math>, a subset <math>A</math> of <math>X</math>. Two distinct points <math>x_1,x_2 \in A</math>.
+'''To prove''': There exist disjoint open subsets <math>U_1,U_2</math> of <math>A</math> such that <math>x_1 \in U_1,x_2 \in U_2</math>.
+'''Proof''':
+{| class="sortable" border="1"
+! Step no. !! Assertion/construction !! Facts used !! Given data used!! Previous steps used !! Explanation
+|-
+| 1 || <math>x_1, x_2</math> are distinct points of <math>X</math> || || <math>x_1,x_2</math> are distinct points of <math>A</math><br><math>A \subseteq X</math> || ||
+|-
+| 2 || There exist disjoint open subsets <math>V_1, V_2</math> of <math>X</math> such that <math>x_1 \in V_1, x_2 \in V_2</math>. || || <math>X</math> is Hausdorff || Step (1) || Step-given  direct
+|-
+| 3 || Define <math>U_1 = V_1 \cap A</math> and <math>U_2 = V_2 \cap A</math>. || || || ||
+|-
+| 4 || <math>U_1, U_2</math> are open subsets of <math>A</math>. || definition of subspace topology || || Steps (2), (3) || By Step (2), <math>V_1,V_2</math> are open, so by the definition of subspace topology, <math>U_1, U_2</math> are open as per their definitions in Step (3).
+|-
+| 5 || <math>U_1, U_2</math> are disjoint. || || || Steps (2), (3) || follows directly from <math>V_1,V_2</math> being disjoint
+|-
+| 6 || <math>x_1 \in U_1, x_2 \in U_2</math> || || <math>x_1,x_2 \in A</math> || Steps (2), (3) || By Step (3), <math>U_1 = V_1 \cap A</math>. By Step (2), <math>x_1 \in V_1</math>, and we are also given that <math>x_1 \in A</math>, so <math>x_1 \in V_1 \cap A = U_1</math>. Similarly, <math>x_2 \in U_2</math>.
+|-
+| 7 || <math>U_1,U_2</math> are the desired open subsets of <math>A</math>. || || || Steps (4)-(6) || Step-combination direct, it's what we want to prove.
+|}
+{{tabular proof format}}
+==References==
-The proof has the following key steps:
+===Textbook references===
-* Start with two points in the subspace
+* {{booklink-proved|Munkres}}, Page 100, Theorem 17.11, Page 101, Exercise 12 and Page 196 (Theorem 31.2 (a))
-* View them as points in the whole space
-* Separate them by disjoint open sets in the whole space
-* Intersect these open sets with the subspace, and use the definition of subspace topology to note that we get disjoint open sets in the subspace separating the points