1101: For Topological Space Induced by Metric and Subset, Subset as Topological Subspace Equals Subset as Topological Space Induced by Metric Subspace

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description/proof of that for topological space induced by metric and subset, subset as topological subspace equals subset as topological space induced by metric subspace

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Topics

About: metric space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of topology induced by metric.
• The reader knows a definition of topological subspace.
• The reader knows a definition of metric subspace.

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Target Context

• The reader will have a description and a proof of the proposition that for any topological space induced by any metric and any subset, the subset as the topological subspace equals the subset as the topological space induced by the metric subspace.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$T^{\prime }$: $\in \{\text{ the topological spaces }\}$ induced by any metric, $dist:T^{\prime }\times T^{\prime }\to \mathbb{R}$
$S$: $\subseteq T^{\prime }$
$T_1$: $=S$ as the topological subspace of $T^{\prime }$
$T_2$: $=S$ as the topological space induced by the metric subspace of $T^{\prime }$
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Statements:
$T_1=T_2$
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2: Proof

Whole Strategy: Step 1: take any open subset of $T_1$, $U\subseteq T_1$, and see that $U$ is an open subset of $T_2$; Step 2: take any open subset of $T_2$, $U\subseteq T_2$, and see that $U$ is an open subset of $T_1$.

Step 1:

Let $U\subseteq T_1$ be any open subset of $T_1$.

$U=U^{\prime }\cap S$ where $U^{\prime }\subseteq T^{\prime }$ is an open subset of $T^{\prime }$, by the definition of subspace topology.

Let $u\in U$ be any.

$u\in U^{\prime }$ and there is an open ball around $u$, $B_{u,ϵ}^{\prime }\subseteq T^{\prime }$, such that $B_{u,ϵ}^{\prime }\subseteq U^{\prime }$, by the definition of topology induced by metric.

$B_{u,ϵ}:=B_{u,ϵ}^{\prime }\cap S\subseteq T_2$ is an open ball in the metric subspace, $T_2$, by the definition of metric subspace.

$B_{u,ϵ}\subseteq U^{\prime }\cap S=U$.

As $u\in U$ is arbitrary, $U$ is an open subset of $T_2$, by the definition of topology induced by metric.

Step 2:

Let $U\subseteq T_2$ be any open subset of $T_2$.

Let $u\in U$ be any.

There is an open ball around $u$, $B_{u,{ϵ}_u}\subseteq T_2$, such that $B_{u,{ϵ}_u}\subseteq U$, where ${ϵ}_u$ means that it depends on $u$, by the definition of topology induced by metric.

$B_{u,{ϵ}_u}=B_{u,{ϵ}_u}^{\prime }\cap S$, by the definition of metric subspace.

Let us take $U^{\prime }:={\cup }_{u\in U}{B}_{u,{ϵ}_u}^{\prime }\subseteq T^{\prime }$, which is an open subset of $T^{\prime }$, by the definition of topology induced by metric.

$U=U^{\prime }\cap S$, because for each $u\in U$, $u\in U^{\prime }\cap S$; for each $u^{\prime }\in U^{\prime }\cap S$, $u^{\prime }\in B_{u,{ϵ}_u}^{\prime }$ and $u^{\prime }\in S$, so, $u^{\prime }\in B_{u,{ϵ}_u}^{\prime }\cap S=B_{u,{ϵ}_u}\subseteq U$.

So, $U\subseteq T_1$ is an open subset of $T_1$, by the definition of subspace topology.

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References

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