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A fixed-point of a function f : A → A is a point a ∈ A such that f(a) = a. The diagonal of A × A is the set of all pairs (a, a) in A × A. (a) Show that f : A → A has a fixed-point if and only if the graph of f intersects the diagonal. (b) Prove that every continuous function f : [0, 1] → [0, 1] has at least one fixed-point. (c) Is the same true for continuous functions f : (0, 1) → (0, 1)?† (d) Is the same true for discontinuous functions?

Question

A fixed-point of a function f : A → A is a point a ∈ A such that f(a) = a. The diagonal of A × A is the set of all pairs (a, a) in A × A. (a) Show that f : A → A has a fixed-point if and only if the graph of f intersects the diagonal. (b) Prove that every continuous function f : [0, 1] → [0, 1] has at least one fixed-point. (c) Is the same true for continuous functions f : (0, 1) → (0, 1)?† (d) Is the same true for discontinuous functions?

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Solution

(a) If f : A → A has a fixed point, then there exists some a ∈ A such that f(a) = a. This can be represented as the point (a, a) which lies on the diagonal of A × A. Therefore, the graph of f intersects the diagonal.

Conversely, if the graph of f intersects the diagonal, then there exists some point (a, a) that lies on both the graph of f and the diagonal. This means that f(a) = a, so a is a fixed point of f.

(b) The function f : [0, 1] → [0, 1] is continuous on a closed interval, so by the Brouwer fixed-point theorem, it has at least one fixed point.

(c) The same is not necessarily true for continuous functions f : (0, 1) → (0, 1). Consider the function f(x) = x/2. This function is continuous on (0, 1), but it does not have a fixed point in this interval.

(d) The same is not necessarily true for discontinuous functions. Consider the function f(x) = 1 if x ≠ 0 and f(0) = 0. This function is discontinuous at x = 0, but it has a fixed point at x = 0. However, not all discontinuous functions will have a fixed point. For example, the function f(x) = 1 if x ≠ 0 and f(0) = 1 does not have a fixed point.

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