Hint for 3.5.7

Suppose f is a monotone increasing function on [a,b], and c is in (a,b).

Let s = sup f( [a,c) ). Prove that if x_n -> c^-, then f(x_n) -> s.

Proof:

Let e > 0.

There exists y in f( [a, c) ) such that

| y - s | < e, by previous HW (2.5.3).

Now, y = f(d) for some d in [a,c).

So | f(d) - s | < e.

Since x_n -> c^-,

there exists N such that for all n > N,

|x_n - c | < c - d,

i.e., x_n is in (d,c].

So, for all n > N, x_n > d,

so f(x_n) >= f(d) since f is monotone increasing;

so | f(x_n) - s | < e because

f(x_n) is in the interval [f(d, s].

So f(x_n) -> s.