Regularity is hereditary: Difference between revisions

Latest revision as of 00:11, 25 January 2012

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

Fill this in later

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:

Opposite facts

Normality is not hereditary

Facts used

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] We are given that $x \in A$ , so if $x \in D$ , then $x \in D \cap A$ . By Step (1), $D \cap A = C$ . Thus, if $x \in D$ , then $x \in C$ , which we know is not true. Hence, $x \notin D$ .
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.

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 196, Theorem 31.2(b), Chapter 4, Section 31

@@ Line 1: / Line 1: @@
-{{topospace metaproperty satisfaction}}
+{{topospace metaproperty satisfaction|
+property = regular space|
+metaproperty = subspace-hereditary property of topological spaces}}
-{{basic fact}}
+==Statement==
+Any subset of a [[regular space]] is regular under the [[subspace topology]].
+==Definitions used==
-==Statement==
+{{fillin}}
-===Property-theoretic statement===
+==Related facts==
-The [[property of topological spaces]] of being a [[regular space]] is a [[hereditary property of topological spaces]].
+===Similar facts===
-===Verbal statement===
+{| class="sortable" border="1"
+! 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]]
+|}
-Any subset of a [[regular space]] is regular under the [[subspace topology]].
+Other similar facts:
-==Definitions used==
+* [[Normality is weakly hereditary]]
+* [[Compactness is weakly hereditary]]
-===Regular space===
+===Opposite facts===
-{{further|[[Regular space]]}}
+* [[Normality is not hereditary]]
-===Subspace topology===
+==Facts used==
-{{further|[[Subspace topology]]}}
+# [[uses::T1 is hereditary]]
 ==Proof==
+'''Given''': A topological space <math>X</math>, a subset <math>A</math> of <math>X</math>. <math>X</math> is regular.
+'''To prove''': <math>A</math> is regular.
+'''Proof''': The T1 property for <math>A</math> follows from Fact (1). It thus suffices to show the separation property for <math>A</math>, i.e., that any point and disjoint closed subset in <math>A</math> can be separated by a continuous function. In other words, we want to prove the following.
-===Proof outline===
+'''To prove (specific)''': For any point <math>x \in A</math> and any closed subset <math>C</math> of <math>A</math> such that <math>x \notin A</math>, there exist disjoint open subsets <math>U_1, U_2</math> of <math>A</math> such that <math>x \in U_1</math> and <math>C \subseteq U_2</math>.
-Any subspace of a [[T1 space]] is T1, so we only need to check separation of points and closed sets. We do this as follows:
+'''Proof (specific)''':
-* Pick a point, and a closed set not containing it, in the subspace (the set is closed relative to the subspace)
+{| class="sortable" border="1"
-* By the definition of [[subspace topology]], find a closed set in the whole space, whose intersection with the subspace is the given closed set)
+! Step no. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation
-* Separate the point and this bigger closed set, in the whole space, by disjoint open sets (using regularity of the whole space)
+|-
-* Intersect these open sets with the subspace to get a separation by disjoint open sets in the subspace
+| 1 || There exists a closed subset <matH>D</math> of <math>X</math> such that <math>C= D \cap A</math>. || definition of subspace topology || <math>C</math> is closed in <math>A</math>. || ||
+|-
+| 2 || <math>x</math> is not in <math>D</math> || || <math>x \in A, x \notin C</math> || Step (1) || <toggledisplay>We are given that <math>x \in A</math>, so if<math>x \in D</math>, then <math>x \in D \cap A</math>. By Step (1), <math>D \cap A = C</math>. Thus, if <math>x \in D</math>, then <math>x \in C</math>, which we know is not true. Hence, <math>x \notin D</math>.</toggledisplay>
+|-
+| 3 || There exist disjoint open subsets <math>V_1, V_2</math> of <math>X</math> such that <math>x \in V_1, D \subseteq V_2</math>. || || <math>X</math> is regular || Steps (1), (2) || Step-given combination direct
+|-
+| 4 || Define <math>U_1 = V_1 \cap A</math> and <math>U_2 = V_2 \cap A</math>. || || || ||
+|-
+| 5 || <math>U_1, U_2</math> are open subsets of <math>A</math>. || definition of subspace topology || || Steps (3), (4) || 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).
+|-
+| 6 || <math>U_1, U_2</math> are disjoint. || || || Steps (3), (4) || follows directly from <math>V_1,V_2</math> being disjoint
+|-
+| 7 || <math>x \in U_1, C \subseteq U_2</math> || || <math>x \in A, C \subseteq A</math> || Steps (3), (4) || By Step (4), <math>U_1 = V_1 \cap A</math>. By Step (3), <math>x \in V_1</math>, and we are also given that <math>x\in A</math>, so <math>x_1 \in V_1 \cap A = U_1</math>. Similarly, <math>C \subseteq V_2</math> and <math>C \subseteq A</math>, so <math>C \subseteq V_2 \cap A = U_2</math>.
+|-
+| 8 || <math>U_1, U_2</math> are the desired open subsets. || || || Steps (5)-(7) || Step-combination, this is exactly what we want to prove.
+|}
-Note that this proof does ''not'' work for normality because we would need to enlarge both [[closed subset]]s, and the process of enlarging might lead to intersection.
+{{tabular proof format}}
 ==References==
 ===Textbook references===
-* {{booklink|Munkres}}, Page 196 (Theorem 31.2(b))
+* {{booklink-proved|Munkres}}, Page 196, Theorem 31.2(b), Chapter 4, Section 31