Topology, Tietze's Extension Theorem

Tietze's Extension Theorem

First assume f is bounded. Adjust the range, so that f(A) fits within (0,1).

Let B be the set of all x in A with f(x) ≤ -1/3, and let C be the set of all x in A with f(x) ≥ 1/3. Since B and C are closed in A, and A is closed, B and C are closed in S. Using Urysohn's lemma, there is a continuous function h₁ that is -1/3 on B and 1/3 on C. If you review the construction of this function, you'll note that h₁ attains its minimum on B, and its maximum on C. Thus -1/3 ≤ h₁(S) ≤ 1/3.

Select an x in A, and it is in B, on its way from B to C, or in C. In each case show that |f(x)-h₁(x)| < 2/3. Call this difference j₁, and apply the above procedure, substituting j₁ for f. This time the image of B is bounded above by -2/9, and the image of C is bounded below by 2/9. This gives a function h₂ with h₂(S) trapped between -2/9 and 2/9. let j₂ = j₁-h₂, and |j₂| < 4/9. Repeat this process, building an infinite sequence of functions.

Let g(x) be the infinite sum over h_i(x). Since h_i is bounded by a geometric series, g(x) is well defined. Also, each h_i is continuous. Use the weierstrass m test to show that convergence is uniform, whence g(S) is continuous.

Remember that h₁ is the first approximation to f, and it is off by j₁. In the same way, h₂ is an approximation to j₁, and it is off by j₂. So h₁+h₂ is an approximation to f, that is off by j₂. Continue this process and j_n is the error term. This approaches 0, hence g(A) = f(A). Thus g is an extension of f.

With |h₁| ≤ 1/3, and |h₂| ≤ 2/9, and |h₃| ≤ 4/27, and so on, evaluate the geometric sum and show that |g| ≤ 1. The bound of g is the bound of f with the endpoints brought in. Let Z be the preimage of ±1 in S. Z and A are disjoint closed sets, and can be separated in disjoint open sets. The topology of Z union A, as a closed set in S, is the topology of Z, and the topology of A, as separate spaces. Define a new function f′ that agrees with f on A, and is equal to 0 on Z. Extend this to a function g′ across all of S. This may also have points that map to 1 or -1, but they will not belong to A or Z. Take the average of g and g′. This is continuous, and agrees with f, and its image lies in (0,1).

Now we can generalize the theorem to an unbounded function f. Set e = f/(1+|f|), so that |e| < 1. Extend e to a function k across all of S as shown above. Then let g = k/(1-|k|). Note that g is continuous, and agrees with f on A.

If A is an open set the theorem fails. Let S be the real line and let A be the complement of 0. Let f(x) = sin(1/x). The image is bounded and continuous, but cannot be extended to x = 0.