description/proof of that metric space is compact iff each infinite subset has \(\omega\)-accumulation point
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of metric space.
- The reader knows a definition of \(\omega\)-accumulation point of subset of topological space.
- 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 with the induced topology 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 for any metric space and any subset, the closure of the subset is compact if and only if each sequence into the subset has a subsequence that converges in the closure of the subset if and only if the closure of the subset is complete and for each positive real number, there is a set of some finite open balls of the-number-radius that covers the closure of the subset.
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.
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.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(M\): \(\in \{\text{ the metric spaces }\}\), with the topology induced by the metric
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Statements:
\(M \in \{\text{ the compact topological spaces }\}\)
\(\iff\)
\(\forall S \in \{\text{ the infinite subsets of } M\} (\exists m \in \{\text{ the } \omega \text{ -accumulation points of } S\})\)
//
2: Note
"\(\omega\)-accumulation point" in the proposition can be replaced with 'accumulation point', because any metric space is a \(T_1\) space, by the proposition that any metric space with the topology induced by the metric is a \(T_1\) topological 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.
3: Proof
Whole Strategy: apply the proposition that any topological space is countably compact if and only if each infinite subset has an \(\omega\)-accumulation point, the proposition that any metric space with the induced topology is 1st-countable, the proposition that any 1st-countable topological space is sequentially compact if the space is countably compact, the proposition that for any metric space and any subset, the closure of the subset is compact if and only if each sequence into the subset has a subsequence that converges in the closure of the subset if and only if the closure of the subset is complete and for each positive real number, there is a set of some finite open balls of the-number-radius that covers the closure of the subset, and the proposition that any topological space is countably compact if and only if each infinite subset has an \(\omega\)-accumulation point; Step 1: suppose that each \(S\) has an \(m\); Step 2: see that \(M\) is compact; Step 3: suppose that \(M\) is compact; Step 4: see that each \(S\) has an \(m\).
Step 1:
Let us suppose that \(S\) has an \(m\).
Step 2:
\(M\) 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.
\(M\) is 1st-countable, by the proposition that any metric space with the induced topology is 1st-countable.
\(M\) is sequentially compact, by the proposition that any 1st-countable topological space is sequentially compact if the space is countably compact.
\(M\) is compact, by the immediate corollary mentioned in the proposition that for any metric space and any subset, the closure of the subset is compact if and only if each sequence into the subset has a subsequence that converges in the closure of the subset if and only if the closure of the subset is complete and for each positive real number, there is a set of some finite open balls of the-number-radius that covers the closure of the subset.
Step 3:
Let us suppose that \(M\) is compact.
Step 4:
\(M\) is countable compact, obviously.
Each infinite subset, \(S\), has an \(\omega\)-accumulation point, \(m \in M\), by the proposition that any topological space is countably compact if and only if each infinite subset has an \(\omega\)-accumulation point.
