A.1 Distance to a subset. Let (E, d) be a metric space, and let A be a subset of E. For x ∈ E, let d(x, A) = inf_{a∈A} d(x, a).

(1) Show that x ↦ d(x, A) is continuous;

Proof. We will actually prove that |d(x, A)−d(y, A)| ≤ d(x, y) for all x, y ∈ E. By symmetry, it suffices to prove that d(x, A) − d(y, A) ≤ d(x, y). Choose ϵ > 0 and let a ∈ A be such that d(y, A) ≤ d(y, a) ≤ d(y, A) + ϵ. We have d(x, A) ≤ d(x, a) and by the triangle inequality, d(x, a) ≤ d(x, y) + d(y, a) so that d(x, A) ≤ d(x, y) + d(y, A) + ϵ which implies that d(x, A) − d(y, A) ≤ d(x, y) + ϵ. Since this is true for all ϵ > 0, we must have d(x, A) − d(y, A) ≤ d(x, y) which is what we wanted to prove. □

(2) Show that d(x, A) = 0 if and only if x ∈ \overline{A}.

Proof. We have d(x, A) = 0 if and only if there exists a sequence {a_{n}}_{n≥1} of A such that d(x, a_{n}) → 0 or in other words, if and only if and only if there exists a sequence {a_{n}}_{n≥1} of A with a_{n} → x that is, if and only if x ∈ \overline{A}. □

A.2 Extrema on compact spaces. Show that if E is a compact metric space and f : E → ℝ is a continuous map, then f admits a maximum on E. That is, there exists x ∈ E such that f(x) ≥ f(y) for any y ∈ E.

Proof. Since E is compact, f(E) ⊂ ℝ is the continuous image of a compact set and it is also compact. The compact subsets of ℝ are the closed and bounded subsets. Since f(E) is bounded, it admits a sup μ = f(x) ∈ ℝ and there exists a sequence μ_{n} ∈ f(E) that converges to it. Since f(E) is closed, the limit of this sequence is also in f(E). Therefore, we have constructed μ ∈ f(E) such that f(y) ≤ f(x) = μ for all y ∈ E, meaning that μ is the maximum of f on E. □

Use this to show that if F_{1}, F_{2} are two disjoint closed subsets of E then there exists μ > 0 such that d(a_{1}, a_{2}) ≥ μ whenever a_{1} ∈ F_{1} and a_{2} ∈ F_{2}.

Proof. Let g : F_{1} ⟶ ℝ be given by g(x) = d(x, F_{2}) for x ∈ F_{1}. By A.1.1, g is a continuous map. Furthermore, F_{1} is compact, and so g(F_{2}) is a compact subset of ℝ. By the same argumemt as above, it must attain a minimum value at some y ∈ F_{1}. Let μ = g(y). We then have that d(a_{1}, a_{2}) ≥ μ for a_{1} ∈ F_{1}, a_{2} ∈ F_{2}. It remains to show that μ > 0. Consider h : F_{2} ⟶ ℝ given by h(z) = d(y, z) for x ∈ F_{2}. h is continuous and F_{2} is compact, so as before, h must attain a minimum at some w ∈ F_{2}, i.e., μ = d(w, y). Since F_{1} and F_{2} are disjoint, y ≠ w and so μ > 0. □

A.3 Contracting maps. Let (E, d) be a compact metric space and let f : E → E be a continuous map such that d(f(x), f(y)) < d(x, y) for all x, y ∈ E with x ≠ y. Define the diameter of a closed subset F of E to be diam(F) = sup_{x,y∈F} d(x, y).

(0) What are the closed subsets of E whose diameter is 0?

Proof. They are the closed subsets made of only one point. □

(1) Show that there exist x, y ∈ F such that diam(F) = d(x, y).

Proof. Since F is closed, it is compact in E and F × F is compact in E × E so d, which is a continuous function on F × F, attains its maximum. Thus sup_{a,b∈F} d(a, b) = d(x, y) for some x, y ∈ F. □

(2) Show that diam(f(F)) < diam(F) if diam(F) ≠ 0;

Proof. By A.3.1 applied to f(F) which is also compact (being the continuous image of a compact set), we see that diam(f(F)) = d(f(w), f(z)) for some w, z ∈ F and so d(w, z) > diam(f(F)) = d(f(w), f(z)). This means that there are two points in F further apart than diam(f(F)): more precisely, diam(F) ≥ d(w, z) > diam(f(F)) = d(f(w), f(z)). □

(3) Use this to show that f has a unique fixed point.

Proof. We will first prove the uniqueness of the fixed point in f. Suppose we have 2 distinct fixed points, x and y. Then d(x, y) = d(f(x), f(y)) < d(x, y), but this is a contradiction.

Now, we prove existence: choose x_{0} ∈ E and define a sequence x_{n} by x_{n+1} = f(x_{n}). We will prove that x_{n} converges to the fixed point of f. Let A be the set of accumulation points of x_{n}. We’ve seen in the previous homework that A is closed. Furthermore, f(A) ⊂ A because if x_{ϕ(n)} → a ∈ A then x_{ϕ(n)+1} = f(x_{ϕ(n)}) → f(a) ∈ A. Finally f : A → A is surjective: if x_{ϕ(n)} → a ∈ A then the sequence x_{ϕ(n)−1} has an accumulation point b and it’s not hard to see that b ∈ A and that f(b) = a. Therefore f(A) = A which contradicts (2) unless A is a point. This implies that x_{n} has a unique accumulation point, i.e. a limit, which is easily seen to be the fixed point of f. □

A.4 Accumulation points II. Let E be a compact metric space, let u = {u_{n}}_{n} be a sequence of elements of E, and let A(u) be the set of its accumulation points. Show that if d(u_{n}, u_{n+1}) → 0 as n → +∞ then A(u) is connected.

Hint: if you can write a space X as the disjoint union of two open sets, then those open sets are also closed.

Proof. We prove this by contradiction. Suppose A(u) is the disjoint union of two open sets F_{1}, F_{2}. Then F_{1}, F_{2} are also closed because F_{1} is the complement of the open set F_{2} and vice-versa. Then F_{1} and F_{2} are also compact, being closed in E.

By A.2.2, there exists μ > 0 such that for a_{1} ∈ F_{1}, a_{2} ∈ F_{2}, d(a_{1}, a_{2}) ≥ μ. If d(u_{n}, u_{n+1}) → 0 as n → +∞ then there exists N such that if n ≥ N then d(u_{n}, u_{n+1}) < μ. This implies that all the u_{n}’s with n ≥ N are in F_{1} or that all of them are in F_{2}, and this in turn implies that A(u) = F_{1} or that A(u) = F_{2}. Hence one of F_{1}, F_{2} is necessarily empty and so A(u) is connected. □