description/proof of that metric space is compact iff it is countably compact
Topics
About: metric space
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topology induced by metric.
- The reader knows a definition of compact topological space.
- The reader knows a definition of countably compact topological space.
- 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 it is countably compact.
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
//
Statements:
\(M \in \{\text{ the compact topological spaces }\}\)
\(\iff\)
\(M \in \{\text{ the countably compact topological spaces }\}\)
//
2: Proof
Whole Strategy: apply 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, and 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 1: suppose that \(M\) is compact; Step 2: see that \(M\) is countably compact; Step 3: suppose that \(M\) is countably compact; Step 4: see that \(M\) is compact.
Step 1:
Let us suppose that \(M\) is compact.
Step 2:
\(M\) is countably compact, because for any countable open cover of \(M\), it is an open cover of \(M\), so, there is a finite subcover, because \(M\) is compact.
Step 3:
Let us suppose that \(M\) is countably compact.
Step 4:
\(M\) 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.
