402: Metric Space Is Compact iff Each Infinite Subset Has \omega-Accumulation Point

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

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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: Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of metric space.
• The reader knows a definition of $\omega$-accumulation point of subset.
• The reader admits the proposition that any topological space is countably compact if and only if each infinite subset has an $\omega$-accumulation point.
• The reader admits the proposition that any metric space is 1st-countable.
• The reader admits the proposition that any 1st-countable topological space is sequentially compact if the space is countably compact.
• The reader admits the proposition that any metric space is compact if and only if it is sequentially compact.

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Target Context

• The reader will have a description and a proof of the proposition that any metric space is 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 metric space, $T$, is 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 $S$ has a $p$. $T$ is countably compact, by the proposition that any topological space is countably compact if and only if each infinite subset has an $\omega$-accumulation point. $T$ is compact, by the proposition that any metric space is 1st-countable, the proposition that any 1st-countable topological space is sequentially compact if the space is countably compact, and the proposition that any metric space is compact if and only if it is sequentially compact.

Let us suppose that $T$ is compact. $T$ is countable compact. Each infinite subset, $S$, has an $\omega$-accumulation point, $p\in T$, by 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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3: Note

"$\omega$-accumulation point" in the proposition can be replaced with accumulation point, because any metric space is a $T_1$ space, while the proposition that on any $T_1$ topological space, any point is an $\omega$-accumulation point of any subset if and only if the point is an accumulation point of the subset holds.

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References

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