349: Quotient Space of Compact Topological Space Is Compact

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A description/proof of that quotient space of compact topological space is compact

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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: Description
• 2: Proof

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Starting Context

• The reader knows a definition of quotient topology on set with respect to map.
• The reader admits the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.

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Target Context

• The reader will have a description and a proof of the proposition that any quotient space of any compact topological space is compact.

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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: Description

For any compact topological space, $T_1$, any quotient space, $T_2$, such that $f:T_1\to T_2$ where $f$ is a quotient map, is compact.

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2: Proof

Let $\{U_{\alpha }\}$ be any open cover of $T_2$ such that ${\cup }_{\alpha }{U}_{\alpha }=T_2$. Each $f^{-1}(U_{\alpha })$ is open on $T_1$ by the definition of quotient topology. $\{f^{-1}(U_{\alpha })\}$ covers $T_1$, because $f^{-1}(T_2)=T_1=f^{-1}({\cup }_{\alpha }{U}_{\alpha })={\cup }_{\alpha }{f}^{-1}(U_{\alpha })$, by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets. So, $\{f^{-1}(U_{\alpha })\}$ is an open cover of $T_1$. As $T_1$ is compact, there is a finite subcover, $\{f^{-1}(U_i)\}$. Is $\{U_i\}$ a cover of $T_2$? Let us suppose that there was a $p\in T_2$ such that $p\notin U_i$ for each $i$. Then, $f^{-1}(p)\cap f^{-1}(U_i)=\mathrm{\varnothing }$, because if $p^{\prime }\in f^{-1}(p)\cap f^{-1}(U_i)$, $f(p^{\prime })=p$ and $f(p^{\prime })\in U_i$, which would mean $p\in U_i$, a contradiction. But as $f$ is surjective, there would be $p^{\prime }\in f^{-1}(p)$ and $p^{\prime }\notin {\cup }_i{f}^{-1}(U_i)$, so, $\{f^{-1}(U_i)\}$ would not be any open cover of $T_1$, a contradiction. So, $\{U_i\}$ is a finite subcover of $\{U_{\alpha }\}$. As there is a finite subcover for any open cover, $T_2$ is compact.

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References

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