A description/proof of that on 2nd-countable topological space, open cover has countable subcover
Topics
About: topological space
The table of contents of this article
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.
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.
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: Description
For any 2nd-countable topological space, \(T\), and any subset, \(S \subseteq T\), any open cover, \(O := \{U_\alpha\vert \alpha \in A\}\), where \(A\) is any possibly uncountable indices set, such that \(S \subseteq \cup_{\alpha \in A} U_\alpha\), has a countable subcover, \(O' := \{U_i\vert i \in I\}\), where \(I \subseteq A\) is a countable indices set, such that \(S \subseteq \cup_{i \in I} U_i\).
2: Proof
Let us denote a countable basis as \(\{B_i\vert i \in I'\}\), where \(I'\) is a countable indices set. Let us define the indices set, \(I'' := \{i \in I'\vert \exists \alpha \in A \owns B_i \subseteq U_\alpha\}\), and the function, \(f: I'' \rightarrow Pow O, i \mapsto \{U_\alpha \in O\vert B_i \subseteq U_\alpha\}\), 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 \(c\). \(c \circ f: I'' \rightarrow O\) 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, \(O' := \{c \circ f (i)\vert i \in I''\}\) is a countable subcover of \(O\).
3: Note
So, any 2nd-countable topological space is compact if and only if it is countably compact.
