A description/proof of that for topological space, open and closed subset of space is union of connected components of space
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of closed set.
- The reader knows a definition of connected topological component.
- The reader admits the proposition that the connected components of any topological space is a partition of the topological space.
- The reader admits the proposition that any topological space is connected if and only if its open and closed subsets are only the topological space and the empty set.
- The reader admits the proposition that for any continuous map between any topological spaces, the image of any connected subspace of the domain is connected on the codomain.
- The reader admits the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space, any open and closed subset is the union of some connected components of the topological space.
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 topological space, \(T\), any open and closed subset, \(S \subseteq T\), is the union of some connected components of \(T\).
2: Proof
If \(S = T\) or \(S = \emptyset\), the proposition is true, because any topological space is the union of the connected components, by the proposition that the connected components of any topological space is a partition of the topological space.
Let us suppose that \(S\) is any nonempty proper subset. \(T\) is not connected, by the proposition that any topological space is connected if and only if its open and closed subsets are only the topological space and the empty set.
Let us think \(S\) as the topological subspace of \(T\). \(S\) is the union of the connected components of \(S\), \(\{C_\alpha\vert \alpha \in A\}\), by the proposition that the connected components of any topological space is a partition of the topological space. The issue is, is any connected component of \(S\) a connected component of \(T\)? Let us think of the inclusion, \(\iota: S \rightarrow T\). As \(\iota\) is continuous, the image of any connected component, \(\iota (C_\alpha) = C_\alpha\) is connected on \(T\), by the proposition that for any continuous map between any topological spaces, the image of any connected subspace of the domain is connected on the codomain. So, \(T\) has the equal or a larger corresponding connected component, \(C'_\alpha\), such that \(C_\alpha \subseteq C'_\alpha\), but \(C'_\alpha \subseteq S\), because otherwise, \(C'_\alpha \cap S\) and \(C'_\alpha \cap (T \setminus S)\) would be a disjoint pair of nonempty open subsets of the subspace, \(C'_\alpha\), whose union would be \(C'_\alpha\), a contradiction against \(C'_\alpha\)'s being a connected subspace. But \(C'_\alpha\) is connected on \(S\), because if \(C'_\alpha = U_1 \cup U_2\) where \(U_i \subseteq S\) is nonempty open on \(S\), \(U_i\) is open on \(T\), by the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space, and if \(U_1 \cap U_2 = \emptyset\), \(C'_\alpha\) would be disconnected on \(T\), a contradiction. So, \(C_\alpha = C'_\alpha\).
