Question

Prove the following (somewhat harder) assertions. If X is nonempty, compact, Hausdorff, and f: X→X is continuous, then there exists A⊂Xsuch that A≠∅, A≠X, and f(A) =A. You can proceed to prove this assertion.

Answer 1

To prove the assertion, we define a set A as the set of all points x in X such that f(x) = x. By utilizing the properties of compactness, continuity, and Hausdorffness, we can show that A is a nonempty, non-X set satisfying f(A) = A.

Step-by-step explanation:

To prove the given assertion, let's start by defining the set A as the set of all points x in X such that f(x) = x. In other words, A = x ∈ X .

Now, since X is compact and f is continuous on X, we know that f(X) is also compact. And since X is Hausdorff, it is a closed subset of X.

Therefore, A = f(X) ∩ X is a nonempty intersection of two compact sets, and is thus compact. Additionally, A is also a closed subset of X since it is the intersection of two closed sets. Hence, A satisfies the conditions A ≠ ∅, A ≠ X, and f(A) = A, as required.