1052: For Continuous Map Between Topological Spaces and Subset of Domain Mapped into Open Subset of Codomain, There Is Open Neighborhood of Domain Subset Mapped into Open Subset

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description/proof of that for continuous map between topological spaces and subset of domain mapped into open subset of codomain, there is open neighborhood of domain subset mapped into open subset

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Topics

About: topological 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 continuous map.

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Target Context

• The reader will have a description and a proof of the proposition that for any continuous map between any topological spaces and any subset of the domain that is mapped into any open subset of the codomain, there is an open neighborhood of the domain subset mapped into the codomain open subset.

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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:
$T_1$: $\in \{\text{ the topological spaces }\}$
$T_2$: $\in \{\text{ the topological spaces }\}$
$f$: $:T_1\to T_2$, $\in \{\text{ the continuous maps }\}$
$S_1$: $\subseteq T_1$
$U_2$: $\in \{\text{ the open subsets of }{T}_2\}$
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Statements:
$f(S_1)\subseteq U_2$
$⟹$
$\mathrm{\exists }{U}_1\subseteq T_1\in \{\text{ the open subsets of }{T}_1\}(S_1\subseteq U_1\wedge f(U_1)\subseteq U_2)$
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2: Proof

Whole Strategy: Step 1: for each $s\in S_1$, take an open neighborhood of $s$, $U_s\subseteq T_1$, such that $f(U_s)\subseteq U_2$; Step 2: take $U_1:={\cup }_{s\in S_1}{U}_s$ and see that $f(U_1)\subseteq U_2$.

Step 1:

Let $s\in S_1$ be any.

$f(s)\in U_2$, which means that $U_2$ is an open neighborhood of $f(s)$.

As $f$ is continuous, there is an open neighborhood of $s$, $U_s\subseteq T_1$, such that $f(U_s)\subseteq U_2$.

Step 2:

Let us define $U_1:={\cup }_{s\in S_1}{U}_s$, which is open on $T_1$.

$S_1\subseteq U_1$, because for each $s\in S_1$, $s\in U_s\subseteq U_1$.

For each $u\in U_1$, $u\in U_s$ for an $s$. So, $f(u)\in U_2$ because $f(U_s)\subseteq U_2$.

So, $f(U_1)\subseteq U_2$.

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References

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