Verified answer

Just to clarify, a Gδ set is an intersection of a countable collection of open sets.

Let (e_n) be a sequence such that e_n = 1/n for all n > 0.

Let d(x,y) be the distance function in M.

For each natural number n, let S_n = {t ∈ T | ∃ a neighbourhood N of t such that d(f(x), f(y)) < e_n for all x,y ∈ N}.

We now have 3 things to prove:

(1) Each S_n is an open set.

Proof: Consider any t ∈ S_n. Let N be the neighbourhood of t described in the definition. By definition of neighbourhood, N must contain an open set O which contains t. Since O ⊆ N, then for all x,y ∈ O, we must have d(f(x), f(y)) < e_n. By definition of open set, O is a neighbourhood of each of its points. Hence, by definition of S_n, every point in O must be in S_n (each point in it has the neighbourhood O such that d(f(x), f(y)) < e_n for all x,y ∈ O).

Thus O ⊆ S_n. Therefore S_n contains a neighbourhood of t, therefore it is a neighbourhood of t.

Hence S_n is a neighbourhood of each of its points, therefore it is an open set.

(2) Every point in D is in each set S_n.

Proof: Suppose t ∈ D. Since f is continuous at t, then by definition there is a neighbourhood N of t such that d(f(t), f(x)) < (1/2)e_n and d(f(t), f(y)) < (1/2)e_n for all x,y ∈ N. Therefore, by the Triangle Inequality,

d(f(x), f(y)) <= d(f(t), f(x)) + d(f(t), f(y)) < e_n

Therefore, d(f(x), f(y)) < e_n for all x,y ∈ N. Therefore, t ∈ S_n.

(3) If a point t is in every one of S_n, then t ∈ D.

Proof: Let ε > 0 be given. By the Archimedean Principle, ∃ n > 0 such that n > 1/ε. Therefore, e_n = 1/n < ε. Since t ∈ S_n, there is a neighbourhood N of t such that d(f(x), f(y)) < e_n < ε, for all x,y ∈ N. If we let y = t, we get that d(f(x), f(t)) < ε.

Therefore, by definition, f is continuous at t. Therefore t ∈ D.

Combining (1), (2) and (3) tells us that D is the intersection of all the open sets S_n.

Therefore, by definition, D is a Gδ set.

QED