Question 6.T.10: Let f : D → R be continuous. If D is closed and bounded, the...
Let f : D → \mathbb{R} be continuous. If D is closed and bounded, then f is uniformly continuous.
Suppose that f is continuous on the closed and bounded set D, but not uniformly continuous. We shall show that this leads to a contradiction.
Since f is not uniformly continuous, there is a positive number ε and two sequences, (x_{n}) and (t_{n}), in D such that
|x_{n} − t_{n}| < \frac{1}{n}, |f(x_{n}) − f(t_{n})| ≥ ε. (6.16)
Since D is bounded, so is (x_{n}), and by Theorem 3.14 it has a convergent subsequence (x_{n_{k}} ). Let x be the limit of (x_{n_{k}} ). Since D is closed, x ∈ D.
The corresponding subsequence (t_{n_{k}}) also converges to x, since
|t_{n_{k}} − x| ≤ |t_{n_{k}} − x_{n_{k}} | + |x_{n_{k}} − x|
< \frac{1}{n_{k}} + |x_{n_{k}} − x| → 0.
Now the continuity of f at x allows us to write
f(x_{n_{k}}) → f(x), f(t_{n_{k}}) → f(x).
Hence |f(x_{n_{k}} ) − f(t_{n_{k}})| → 0, which is inconsistent with (6.16)