Difference between revisions of "Subspace topology"

Revision as of 03:12, 22 June 2015

Definition

Given a topological space $(X,\mathcal{J})$ and given a $Y\subset X$ ( $Y$ is a subset of $X$ ) we define the subspace topology as follows:^[1]

$(Y,\mathcal{K})$ is a topological space where the open sets, $\mathcal{K}$ , are given by $\mathcal{K}:=\{Y\cap V\vert\ V\in\mathcal{J}\}$

We may say any one of:

Let $Y$ be a subspace of $X$
Let $Y$ be a subspace of $(X,\mathcal{J})$

and it is taken implicitly to mean $Y$ is considered as a topological space with the subspace topology inherited from $(X,\mathcal{J})$

Proof of claims

[Expand]
Claim 1: The subspace topology is indeed a topology

Here $(X,\mathcal{J})$ is a topological space and $Y\subset X$ and $\mathcal{K}$ is defined as above, we will prove that $(Y,\mathcal{K})$ is a topology.

Recall that to be a topology $(Y,\mathcal{K})$ must have the following properties:

$\emptyset\in\mathcal{K}$ and $Y\in\mathcal{K}$
For any $U,V\in\mathcal{K}$ we must have $U\cap V\in\mathcal{K}$
For an arbitrary family {Uα}α∈I of open sets (that is ∀α∈I[Uα∈K]) we have:
- $\bigcup_{\alpha\in I}A_\alpha\in\mathcal{K}$

Proof:

First we must show that ∅,Y∈K
Recall that ∅,X∈J and notice that:
- $\emptyset\cap Y=\emptyset$ , so by the definition of $\mathcal{K}$ we have $\emptyset\in\mathcal{K}$
- $X\cap Y=Y$ , so by the definition of $\mathcal{K}$ we have $Y\in\mathcal{K}$
Next we must show...

TODO: Easy work just takes time to write!

Terminology

A closed subspace (of $X$ ) is a subset of $X$ which is closed in $X$ and is imbued with the subspace topology
A open subspace (of $X$ ) is a subset of $X$ which is open in $X$ and is imbued with the subspace topology

TODO: Find reference

A set U⊆X is open relative to $Y$ (or relatively open if it is obvious we are talking about a subspace Y of X) if U is open in Y
- This implies that $U\subseteq Y$ ^[1]
A set U⊆X is closed relative to $Y$ (or relatively closed if it is obvious we are talking about a subspace Y of X) if U is closed in Y
- This also implies that $U\subseteq Y$

Immediate theorems

[Expand]
Theorem: Let $Y$ be a subspace of $X$ , if $U$ is open in $Y$ and $Y$ is open in $X$ then $U$ is open in $X$ ^[1]

This may be easier to read symbolically:

if $U\in\mathcal{K}$ and $Y\in\mathcal{J}$ then $U\in\mathcal{J}$

Proof:

Since

$U$ is open in

$Y$ we know that:

$U=Y\cap V$ for some $V$ open in $X$ (for some $V\in\mathcal{J}$ )

Since

$Y$ and

$V$ are both open in

$X$ we know:

$Y\cap V$ is open in $X$

it follows that

$U$ is open in

$X$

References

↑ ^{Jump up to: 1.0} ^1.1 ^1.2 Topology - Second Edition - Munkres

[Topology-1] {Jump up to: 1.0} ^1.1 ^1.2 Topology - Second Edition - Munkres

[1]

@@ Line 1: / Line 1: @@
 ==Definition==
-We define the subspace [[Topological space|topology]] as follows.
+Given a [[Topological space|topological space]] {{M|(X,\mathcal{J})}} and given a {{M|Y\subset X}} ({{M|Y}} is a subset of {{M|X}}) we define the ''subspace topology'' as follows:<ref name="Topology">Topology - Second Edition - Munkres</ref>
+* {{M|(Y,\mathcal{K})}} is a topological space where the [[Open set|open sets]], {{M|\mathcal{K} }}, are given by {{M|1=\mathcal{K}:=\{Y\cap V\vert\ V\in\mathcal{J}\} }}
+We may say any one of:
+# Let {{M|Y}} be a subspace of {{M|X}}
+# Let {{M|Y}} be a subspace of {{M|(X,\mathcal{J})}}
+and it is taken implicitly to mean {{M|Y}} is considered as a topological space with the ''subspace topology'' inherited from {{M|(X,\mathcal{J})}}
-Given a topological space <math>(X,\mathcal{J})</math> and any <math>Y\subset X</math> we can define a topology on <math>Y,\ (Y,\mathcal{J}_Y)</math> where <math>\mathcal{J}_Y=\{Y\cap U|U\in\mathcal{J}\}</math>
+==Proof of claims==
+{{Begin Theorem}}
+Claim 1: The subspace topology is indeed a topology
+{{Begin Proof}}
+Here {{M|(X,\mathcal{J})}} is a topological space and {{M|Y\subset X}} and {{M|\mathcal{K} }} is defined as above, we will prove that {{M|(Y,\mathcal{K})}} is a topology.
-We may say "<math>Y</math> is a subspace of <math>X</math> (or indeed <math>(X,\mathcal{J})</math>" to implicitly mean this topology.
-==Closed subspace==
+Recall that to be a topology {{M|(Y,\mathcal{K})}} must have the following properties:
-If {{m|Y}} is a "closed subspace" of {{m|(X,\mathcal{J})}} then it means that {{M|Y}} is [[Closed set|closed]] in {{M|X}} and should be considered with the subspace topology.
+# {{M|\emptyset\in\mathcal{K} }} and {{M|Y\in\mathcal{K} }}
+# For any {{M|U,V\in\mathcal{K} }} we must have {{M|U\cap V\in\mathcal{K} }}
+# For an [[Indexing set|arbitrary family]] {{M|\{U_\alpha\}_{\alpha\in I} }} of open sets (that is {{M|\forall\alpha\in I[U_\alpha\in\mathcal{K}]}}) we have:
+#* {{MM|\bigcup_{\alpha\in I}A_\alpha\in\mathcal{K} }}
-==Open subspace==
-{{Todo|same as closed, but with the word "open"}}
-==Open sets in open subspaces are open==
+'''Proof:'''
-{{Todo|easy}}
+# First we must show that {{M|\emptyset,Y\in\mathcal{K} }}
+#: Recall that {{M|\emptyset,X\in\mathcal{J} }} and notice that:
+#:* {{M|1=\emptyset\cap Y=\emptyset}}, so by the definition of {{M|\mathcal{K} }} we have {{M|\emptyset\in\mathcal{K} }}
+#:* {{M|1=X\cap Y=Y}}, so by the definition of {{M|\mathcal{K} }} we have {{M|Y\in\mathcal{K} }}
+# Next we must show...
+{{Todo|Easy work just takes time to write!}}
+{{End Proof}}{{End Theorem}}
+==Terminology==
+* A '''closed subspace''' (of {{M|X}}) is a subset of {{M|X}} which is closed in {{M|X}} and is imbued with the subspace topology
+* A '''open subspace''' (of {{M|X}}) is a subset of {{M|X}} which is open in {{M|X}} and is imbued with the subspace topology
+{{Todo|Find reference}}
+* A set {{M|U\subseteq X}} is '''open relative to {{M|Y}}''' (or [[Relatively open|''relatively open'']] if it is obvious we are talking about a subspace {{M|Y}} of {{M|X}}) if {{M|U}} is open in {{M|Y}}
+** This implies that {{M|U\subseteq Y}}<ref name="Topology"/>
+* A set {{M|U\subseteq X}} is '''closed relative to {{M|Y}}''' (or [[Relatively closed|''relatively closed'']] if it is obvious we are talking about a subspace {{M|Y}} of {{M|X}}) if {{M|U}} is [[Closed set|closed]] in {{M|Y}}
+** This also implies that {{M|U\subseteq Y}}
+==Immediate theorems==
+{{Begin Theorem}}
+Theorem: Let {{M|Y}} be a subspace of {{M|X}}, if {{M|U}} is open in {{M|Y}} and {{M|Y}} is open in {{M|X}} then {{M|U}} is open in {{M|X}}<ref name="Topology"/>
+{{Begin Proof}}
+This may be easier to read symbolically:
+* if {{M|U\in\mathcal{K} }} and {{M|Y\in\mathcal{J} }} then {{M|U\in\mathcal{J} }}
+'''Proof:'''
+: Since {{M|U}} is open in {{M|Y}} we know that:
+:* {{M|1=U=Y\cap V}} for some {{M|V}} open in {{M|X}} (for some {{M|V\in\mathcal{J} }})
+: Since {{M|Y}} and {{M|V}} are both open in {{M|X}} we know:
+:* {{M|Y\cap V}} is open in {{M|X}}
+: it follows that {{M|U}} is open in {{M|X}}
+{{End Proof}}{{End Theorem}}
+==References==
+<references/>
 {{Definition|Topology}}

Difference between revisions of "Subspace topology"

Revision as of 03:12, 22 June 2015

Contents

Definition

Proof of claims

Terminology

Immediate theorems

References

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