Let A be a closed subset of a normal space X, and let f: A → [−1, 1] be continuous. What does the Tietze extension theorem guarantee?
AThere exists a continuous F: X → ℝ such that F|_A = f, and F may take values outside [−1, 1] on X \ A
BThere exists a continuous F: X → [−1, 1] such that F|_A = f
Cf extends continuously to X if and only if A is compact
DThere exists a continuous F: X → [−1, 1] such that F|_A = f only when A is also open
The Tietze theorem guarantees both existence of an extension and range control: F can be arranged to map all of X into [−1, 1], not just into ℝ. The correct answer (B) states the stronger bounded version. Option A is also valid (a weaker conclusion also follows), but option B is the most precise and useful statement. Options C and D introduce incorrect conditions: compactness of A is not required, and closedness of A (not openness) is the key hypothesis. Closedness matters because it ensures that boundary behavior of f on A can be matched continuously from outside.
Question 2 Multiple Choice
Why does the proof of the Tietze extension theorem use Urysohn's lemma iteratively, rather than extending f directly?
AUrysohn's lemma is needed to first verify that A is closed before the extension can proceed
BUrysohn's lemma constructs functions that separate 0 and 1 on disjoint closed sets, and these are used as building blocks to approximate f with successively smaller error, assembling the extension as a uniformly convergent series
CUrysohn's lemma replaces f with a simpler linear function that is then extended by linearity
DThe proof uses Urysohn's lemma only once, to separate A from the rest of X
The constructive proof proceeds by building a sequence of continuous functions, each defined on all of X, that approximate f on A with error shrinking by a factor of 2/3 at each step. Each approximation step calls Urysohn's lemma to construct a function that separates two closed sets (where f is large from where f is small). The sum of this series converges uniformly to the desired extension F. Urysohn's lemma is the 'atom' — it provides the simplest kind of separation function — and Tietze shows complex continuous functions are just infinite combinations of these atoms.
Question 3 True / False
The Tietze extension theorem can be viewed as a generalization of Urysohn's lemma, since Urysohn's lemma constructs extensions of specific simple functions (the 0-on-A, 1-on-B indicator) while Tietze handles arbitrary continuous functions.
TTrue
FFalse
Answer: True
Urysohn's lemma solves a special instance of the extension problem: it extends the function that is 0 on the closed set A and 1 on the closed set B to a continuous function on all of X. Tietze generalizes this: instead of the simple 0/1 target, any continuous f: A → ℝ can be extended. The proof of Tietze actually reduces to repeated applications of Urysohn's lemma, confirming the logical dependence. In this sense Urysohn is the atomic building block and Tietze is the general statement.
Question 4 True / False
If A is an open (rather than closed) subset of a normal space X and f: A → ℝ is continuous, then the Tietze theorem still guarantees that f extends to a continuous function on most of X.
TTrue
FFalse
Answer: False
Closedness of A is an essential hypothesis, not a convenience. When A is open, f need not extend continuously: consider X = ℝ, A = (0, 1), and f(x) = 1/x. This is continuous on A but cannot be extended continuously to 0, which is a limit point of A not in A. The open set provides no control over what f does as you approach the boundary from inside, so no continuous extension to the boundary (and hence to all of X) need exist. Closedness ensures that f's values on A constrain what F must do near A's boundary from both sides.
Question 5 Short Answer
Why does the Tietze extension theorem require A to be closed in X, and what can go wrong if A is merely an arbitrary subset?
Think about your answer, then reveal below.
Model answer: Closedness of A ensures that the boundary of A (the points in X that are limits of sequences in A but not in A) is contained in A. This means f is already defined and continuous at every boundary point of A, so there is no ambiguity about what F must do as it approaches A from outside — it must approach the values f takes on A's boundary. If A were not closed, its boundary points would lie outside A, and f would have no values there, creating a gap: F would need to pick values at those boundary points without guidance, and continuity of F on all of X would then be impossible to guarantee. Closedness is what allows the Urysohn-based approximation to 'stitch in' values outside A consistently.
A deeper way to see this: the extension theorem is really a statement about the ability to interpolate continuously from a subset to the whole space. Closed sets are 'complete' in the sense that they contain their own limit points, so the behavior of f on A already determines what F must do at every approach to A from outside. For open sets, you can approach the boundary from outside A without the set 'telling you' what value to take — there is no constraint, and the extension can fail.