A description/proof of that inclusion into topological space from closed subspace is closed continuous embedding
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of subspace topology.
- The reader knows a definition of closed set.
- The reader knows a definition of closed map.
- The reader knows a definition of continuous embedding.
- The reader admits the proposition that for any topological space, the inclusion from any subspace into the topological space is continuous.
- The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
- The reader admits the proposition that any closed set on any closed topological subspace is closed on the base space.
- The reader admits the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset.
- The reader admits the proposition that any topological spaces map is continuous if and only if the preimage of any closed subset of the codomain is closed.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space, the inclusion from any closed subspace is a closed continuous embedding.
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, \(T_1\), and any closed subspace, \(T_2 \subseteq T_1\), the inclusion, \(f: T_2 \rightarrow T_1\) is a closed continuous embedding.
2: Proof
\(f': T_2 \rightarrow f (T_2)\) is a bijection. By the proposition that for any topological space, the inclusion from any subspace is continuous, \(f\) is continuous, and by the proposition that any restriction of any continuous map on the domain and the codomain is continuous, \(f'\) is continuous. For any closed subset, \(C \in T_2\), \({f'^{-1}}^{-1} (C) = f' (C) = C\), which is closed on \(T_1\), by the proposition that any closed set on any closed topological subspace is closed on the base space. \(C = C \cap f (T_2)\) is closed on \(f (T_2)\), by the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset. By the proposition that any topological spaces map is continuous if and only if the preimage of any closed subset of the codomain is closed, \(f'^{-1}\) is continuous. So, \(f'\) is a homeomorphism. \(f'\) is closed because the image of any closed subset of the domain is closed, as has been shown already.
