Inclusion of a subspace is continuous
Builds on Subspace topologyContinuous map of topological spacesPreimage of a set
Statement
Let be a topological space, let have the subspace topology and let be the inclusion map . Then is continuous.
Remarks
Proof: for open in , the preimage is , which is open in . Thus the restriction of a continuous map is continuous. A map is continuous if and only if is continuous. This is the universal property of the subspace topology.
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- Element definition
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- Set definition
- Conjunction definition
- Disjunction definition
- Implication definition
- Predicate definition
- Empty set object
- Family of sets definition
- Ordered pair definition
- Intersection definition
- Union definition
- Existential quantifier definition
- Universal quantifier definition
- Cartesian product definition
- Subset definition
- Union of a family of sets definition
- Uniqueness quantifier definition
- Binary relation definition
- Topological space definition
- Function definition
- Subspace topology definition
- Preimage of a set definition
- Metric space definition
- Open ball definition
- Metric topology definition
- Open set definition
- Continuous map of topological spaces definition
- Inclusion of a subspace is continuous statement
Further reading
About this page
| Tag | 065J822CKN |
|---|---|
| URI | m2://065J822CKN |
| Kind | statement, proposition |
| Writing | W0 stub |
| MSC | 54B05 |
| Status | This copy of the site has no build data. Run m2 status 065J822CKN. |
| Uses | Subspace topology, Continuous map of topological spaces, Preimage of a set |
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