389: On 2nd-Countable Topological Space, Open Cover Has Countable Subcover

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A description/proof of that on 2nd-countable topological space, open cover has countable subcover

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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
• 3: Note

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Starting Context

• The reader knows a definition of 2nd-countable topological space.
• The reader knows a definition of open cover of subset.
• The reader knows a definition of subcover of open cover of subset.

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Target Context

• The reader will have a description and a proof of the proposition that on any 2nd-countable topological space, any open cover of any subset has a countable subcover.

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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 2nd-countable topological space, $T$, and any subset, $S\subseteq T$, any open cover, $O:=\{U_{\alpha }|\alpha \in A\}$, where $A$ is any possibly uncountable indices set, such that $S\subseteq {\cup }_{\alpha \in A}{U}_{\alpha }$, has a countable subcover, $O^{\prime }:=\{U_i|i\in I\}$, where $I\subseteq A$ is a countable indices set, such that $S\subseteq {\cup }_{i\in I}{U}_i$.

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2: Proof

Let us denote a countable basis as $\{B_i|i\in I^{\prime }\}$, where $I^{\prime }$ is a countable indices set. Let us define the indices set, $I^{″}:=\{i\in I^{\prime }|\mathrm{\exists }\alpha \in A\ni B_i\subseteq U_{\alpha }\}$, and the function, $f:I^{″}\to PowO,i\mapsto \{U_{\alpha }\in O|B_i\subseteq U_{\alpha }\}$, which is not necessarily injective, but that is not any problem. With respect to $O$, there is a choice function from the set of the nonempty subsets to $O$ that (the function) chooses an element from each nonempty subset, denoted by $c$. $c\circ f:I^{″}\to O$ is well-defined. The range of $c\circ f$ covers $S$, because for any $p\in S$, there is a $U_{\alpha }\in O$ such that $p\in U_{\alpha }$, there is an $i\in I^{″}$ such that $p\in B_i\subseteq U_{\alpha }$, and there is the $c\circ f(i)$, which may not be $U_{\alpha }$ but contains $B_i$ anyway, so, $p$ is contained in $c\circ f(i)$. So, $O^{\prime }:=\{c\circ f(i)|i\in I^{″}\}$ is a countable subcover of $O$.

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3: Note

So, any 2nd-countable topological space is compact if and only if it is countably compact.

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References

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