# Tietze extension theorem

*This article gives the statement, and possibly proof, of a basic fact in topology.*

## Statement

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 .

## Facts used

- 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.

## Proof

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 . |