390: Topological Space Is Countably Compact iff Each Infinite Subset Has \omega-Accumulation Point

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A description/proof of that topological space is countably compact iff each infinite subset has $\omega$-accumulation point

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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 countably compact topological space.
• The reader knows a definition of $\omega$-accumulation point of subset.

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Target Context

• The reader will have a description and a proof of the proposition that any topological space is countably compact if and only if each infinite subset has an $\omega$-accumulation point.

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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

Any topological space, $T$, is countably compact if and only if each infinite subset, $S$, has an $\omega$-accumulation point, $p\in T$.

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2: Proof

Let us suppose that each $S$ has a $p$. Let us suppose that there was an open countable cover of $T$, $O=\{U_i|i\in I\}$, where $I$ was a countable indices set, without any finite subcover. Let us suppose that $U_i\ne \mathrm{\varnothing }$, because any empty set could be removed and the elements could be re-indexed. Let us suppose that $I=\{1,2,...\}$ without loss of generality. There would be a point, $p_1\in U_1$, and let us define $i_1:=1$ and $U_1^{\prime }:=U_1$. There would be a point, $p_2\notin U_1^{\prime }$, so, $p_2\in U_{i_2}$ for an $i_1<i_2\in I$, and let us define $U_2^{\prime }:={\cup }_{1\le i\le i_2}{U}_i$. There would be a point, $p_3\notin U_2^{\prime }$, so, $p_3\in U_{i_3}$ for an $i_2<i_3\in I$, and so on, which would go on infinitely because there would be no finite subcover of $O$. $O^{\prime }:=\{U_i^{\prime }|i\in I\}$ would be an open cover of $T$. Let us choose $S=\{p_1,p_2,...\}$ and let $p$ be an $\omega$-accumulation point of $S$. $p\in U_j^{\prime }$ for an $j\in I$, because $O^{\prime }$ would be a cover of $T$. But while $U_j^{\prime }$ would be a neighborhood of $p$, $U_j^{\prime }\cap S=\{p_1,p_2,...,p_j\}$, which is a contradiction against $p$'s being an $\omega$-accumulation point.

Let us suppose that $T$ is countably compact. Let us suppose that there was an infinite subset, $S$, that had no $\omega$-accumulation point. Also any countable subset, $S^{\prime }\subseteq S$, would have no $\omega$-accumulation point, because for any point, $p\in T$, there would be a neighborhood, $U_p$, such that $U_p\cap S$ would have only finite points, and also $U_p\cap S^{\prime }$ would have only finite points, because $S^{\prime }$ would be smaller than $S$. For each finite (can be empty) $S^{″}\subseteq S^{\prime }$, let us define $f(S^{″}):=\cup \{U_p|U_p\cap S^{\prime }=S^{″}\}$, which might be empty for an $S^{″}$, but that would not matter. $O:=\{f(S^{″})|S^{″}\subseteq S^{\prime }\}$ would be an open cover of $T$, because as $S^{\prime }$ would have no $\omega$-accumulation point, any point, $p\in T$, would have at least 1 $U_p$ such that $U_p\cap S^{\prime }$ would have only finite points, and in fact, countable, because the set of the finite subsets of any countable set would be countable: for any countable set, $S^{\prime }=\{1,2,...\}$, choose the finite sets in this order: 1st take $\mathrm{\varnothing }$; 2nd take $\{1\}$ and choose $\{1\}$; 3rd take $\{1,2\}$ and choose $\{2\}$ and $\{1,2\}$; 4th take $\{1,2,3\}$ and choose $\{3\}$, $\{2,3\}$, and $\{1,2,3\}$, for example.

There would be a finite subcover, $O^{\prime }:=\{f(S_i^{″})\}\subseteq O$. There would be a point, $p\in S^{\prime }\setminus {\cup }_i{S}_i^{″}$, because $S^{\prime }$ would be infinite while ${\cup }_i{S}_i^{″}$ would be finite. But $p\notin f(S_i^{″})$ for any $i$, because $f(S_i^{″})\cap S^{\prime }=S_i^{″}$, a contradiction against $O^{\prime }$'s being a cover of $T$.

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References

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