Tietze extension theorem

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

Statement

Suppose X 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 A is a closed subset of X, and f:A \to [0,1] is a continuous map. Then, there exists a continuous map g:X \to [0,1] such that the restriction of g to A is f.

Facts used

  1. Urysohn's lemma: This states that for, given two closed subsets A,B of a normal space X, there is a continuous function h:X \to [0,1] such that h(A) = 0 and h(B) = 1.
  2. 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 [0,1] is homeomorphic to [-1,1], it suffices to prove the result replacing [0,1] with [-1,1]. We will also freely use that any closed interval is homeomorphic to [0,1], so Urysohn's lemma can be stated replacing [0,1] by any closed interval.

Given: A normal space X. A closed subset A of X. A continuous function f:A \to [-1,1].

To prove: There exists a continuous function g:X \to [-1,1] such that the restriction of g to A is f.

Proof: We write f_0 = f.

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 \! C_1 = f^{-1}([-1,-1/3]) and \! D_1 = f^{-1}([1/3,1]) Existence of f:A \to [0,1] -- --
2 C_1 and D_1 are disjoint closed subsets of A f is continuous (1) inverse image of closed subset under continuous map is closed C_1 and D_1 are closed subsets of A since they are both the inverse image of a closed set under a continuous map. Moreover, they are disjoint since their images under f are disjoint.
3 C_1 and D_1 are disjoint closed subsets of X A is closed in X (2) (2) By step (2), C_1 and D_1 are closed in A, which is closed in X. Thus, C_1 and D_1 are closed in X.
4 Construct a continuous function \! g_1: X \to [-1/3,1/3] such that \! g_1(C_1) = -1/3 and \! g_1(D_1) = 1/3 (3) (1) (Urysohn's lemma) Urysohn's lemma guarantees a continuous function h:X \to [0,1] such that h(C_1) = 0 and h(D_1) = 1. Define g_1(x) := (2h(x) - 1)/3.
5 Define \! f_1 = f_0 - g_1 as a function on A. Note that \! f_1 is a continuous function on \! A taking values in \! [-2/3,2/3]. (f = f_0 is continuous) (4) sum or difference of continuous functions on subsets of a topological space is continuous on their intersection. f_1 is continuous since both f_0 and g_1 are continuous. For the range, note the following. For x \in A, there are three possibilities for the interval in which f_0(x) lies: [-1,-1/3], [-1/3,1/3], and [1/3,1]. In the first case, g_1(x) = -1/3, so f_0(x) - g_1(x is in [-1 - (-1/3), -1/3 - (-1/3)] = [-2/3,0]. In the second case, both f_0(x) and g_1(x) are in [-1/3,1/3], so by the triangle inequality, the difference is in [-2/3,2/3]. In the third case, g_1(x) = 1/3 so f_0(x) - g_1(x) is in [1/3 - 1/3,1 - 1/3] = [0,2/3].
6 We proceed iteratively as follows: from the previous stage, we have a continuous function f_i on the closed subset A taking values in \! [-(2/3)^i,(2/3)^i].
7 Let \! C_{i+1} = f_i^{-1}[-(2/3)^i,-(1/3)(2/3)^i] and \! D_{i+1} = f_i^{-1}[(1/3)(2/3)^i,(2/3)^i].
8 C_{i+1} and D_{i+1} are disjoint closed subsets of A. (6): f_i is continuous inverse image of closed subset under continuous map is closed C_1 and D_1 are closed subsets of A since they are both the inverse image of a closed set under a continuous map. Moreover, they are disjoint since their images under f are disjoint.
9 C_{i+1} and D_{i+1} are disjoint closed subsets of X A is closed in X (8) (2) (closed subsets of closed subsets are closed) By step (8), C_{i+1} and D_{i+1} are closed in A, which is closed in X. Thus, C_{i+1} and D_{i+1} are closed in X.
10 Find a continuous function g_{i+1}:X \to [-(1/3)(2/3)^i,(1(3/(2/3)^i] such that \! g_{i+1}(C_{i+1}) = (-1/3)(2/3)^i and \! g_{i+1}(D_{i+1}) = (1/3)(2/3)^i. (9) (1) (Urysohn's lemma) We use fact (1) to get a function to [0,1], then multiply by 2 and subtract 1 to get a function to [-1,1], then scale it by the factor of (1/3)(2/3)^i.
11 Define \! f_{i+1} = f_i - g_{i+1}. f_{i+1} is a continuous function on A taking values in \! [-(2/3)^{i+1},(2/3)^{i+1}]. (6), (10) (Same logic as for (5))
12 Define the following function g:X \to [-1,1]: g = \sum_{i=1}^\infty g_i. This function is well defined. (10) Note that the absolute value of g_i is bounded by the geometric progression (1/3) + (1/3)(2/3) + (1/3)(2/3)^2 + \dots = 1. Similarly, the lower bound is -1. Further, since g_n are bounded by the geometric progression in absolute value, the series g_n converges. So the sum is well-defined and takes values in [-1,1].
13 The function g is continuous (10) This follows from the fact that each g_i is continuous and the co-domain of the g_is approaches zero. Fill this in later
14 \! g|_A = f Let x \in A. Then, \! f_1(x) = f(x) - g_1(x). Inductively, f_n(x) = f(x) - \sum_{i=1}^n g_i(x). Since the upper and lower bound on f_n tend to zero as n \to \infty, f_n(x) \to 0 as n \to \infty. Thus, f(x) = \sum_{i=1}^\infty g_i(x) = g(x) for x \in A.