Tietze extension theorem
This article gives the statement, and possibly proof, of a basic fact in topology.
Suppose is a normal space (i.e., a topological space that is T1 and where any two disjoint closed subsets can be separated by disjoint open subsets). Suppose is a closed subset of , and is a continuous map. Then, there exists a continuous map such that the restriction of to is .
- Urysohn's lemma: This states that for, given two closed subsets of a normal space , there is a continuous function such that and .
- Closedness is transitive: A closed subset (in the subspace topology) of a closed subset is closed in the whole space.
This proof uses a tabular format for presentation. Learn more about tabular proof formats|View all pages on facts with proofs in tabular format
Note that since is homeomorphic to , it suffices to prove the result replacing with . We will also freely use that any closed interval is homeomorphic to , so Urysohn's lemma can be stated replacing by any closed interval.
Given: A normal space . A closed subset of . A continuous function .
To prove: There exists a continuous function such that the restriction of to is .
Proof: We write .
The key part of the proof is the inductive portions (6)-(10). Steps (1)-(5) can be thought of as a special case of these; however, these steps are provided separately so that the full details can be seen in the base case before the general inductive setup is used.
|Step no.||Construction/assertion||Given data used||Previous steps used||Facts used||Explanation|
|1||Let and||Existence of||--||--|
|2||and are disjoint closed subsets of||is continuous||(1)||inverse image of closed subset under continuous map is closed||and are closed subsets of since they are both the inverse image of a closed set under a continuous map. Moreover, they are disjoint since their images under are disjoint.|
|3||and are disjoint closed subsets of||is closed in||(2)||(2)||By step (2), and are closed in , which is closed in . Thus, and are closed in .|
|4||Construct a continuous function such that and||(3)||(1) (Urysohn's lemma)||Urysohn's lemma guarantees a continuous function such that and . Define .|
|5||Define as a function on . Note that is a continuous function on taking values in .||( is continuous)||(4)||sum or difference of continuous functions on subsets of a topological space is continuous on their intersection.||is continuous since both and are continuous. For the range, note the following. For , there are three possibilities for the interval in which lies: , , and . In the first case, , so is in . In the second case, both and are in , so by the triangle inequality, the difference is in . In the third case, so is in .|
|6||We proceed iteratively as follows: from the previous stage, we have a continuous function on the closed subset taking values in .|
|7||Let and .|
|8||and are disjoint closed subsets of .||(6): is continuous||inverse image of closed subset under continuous map is closed||and are closed subsets of since they are both the inverse image of a closed set under a continuous map. Moreover, they are disjoint since their images under are disjoint.|
|9||and are disjoint closed subsets of||is closed in||(8)||(2) (closed subsets of closed subsets are closed)||By step (8), and are closed in , which is closed in . Thus, and are closed in .|
|10||Find a continuous function such that and .||(9)||(1) (Urysohn's lemma)||We use fact (1) to get a function to , then multiply by 2 and subtract 1 to get a function to , then scale it by the factor of .|
|11||Define . is a continuous function on taking values in .||(6), (10)||(Same logic as for (5))|
|12||Define the following function : . This function is well defined.||(10)||Note that the absolute value of is bounded by the geometric progression . Similarly, the lower bound is . Further, since are bounded by the geometric progression in absolute value, the series converges. So the sum is well-defined and takes values in .|
|13||The function is continuous||(10)||This follows from the fact that each is continuous and the co-domain of the s approaches zero. Fill this in later|
|14||Let . Then, . Inductively, . Since the upper and lower bound on tend to zero as , as . Thus, for .|