About boundedness of continuous functions on a bounded closed interval Suppose that $f(x)$ is continuous on a bounded closed interval $I$.
Then, $f(x)$ is bounded on $I$.
I want to prove this fact like the following:
Assume that $f(x)$ is not bounded on $I$.
Then, there is an $a \in I$ such that $\displaystyle \lim_{x \to a} f(x) = +\infty$ or $\displaystyle \lim_{x \to a} f(x) = -\infty$.
On the other hand, $\displaystyle \lim_{x \to a} f(x) = f(a)$ because $f(x)$ is continuous at $a \in I$. This is a contradiction.
How can I derive the existence of an $a \in I$ such that $\displaystyle \lim_{x \to a} f(x) = +\infty$ or $\displaystyle \lim_{x \to a} f(x) = -\infty$ ?
 A: If $f$ is unbounded on $I$ it does not necessarily mean that there is a value $a \in I$ for which $f(x) \to \pm\infty$ as $x \to a$. For example consider $f(x) = (1/x)\sin(1/x)$ on $(0, 1)$. It is essential to use some form of completeness here and also note that the the requirement of closed interval is essential. Result fails to hold for functions continuous on open intervals (see example $f$ given earlier).
While there are many way to prove this result (based on many different forms of completeness property) I find the one via Heine Borel Theorem to be very direct. Let $f$ be continuous on $[a, b]$ and defined $f(x) = f(a)$ for $x < a$ and $f(x) = f(b)$ for $x > b$ so that $f$ is continuous everywhere. By continuity for each $x \in [a, b]$ there is an open interval $I_{x}$ containing $x$ such that $|f(t) - f(x)| < 1$ for all $t \in I_{x}$. Thus $f$ is bounded on $I_{x}$ and let $M_{x}$ be such that $|f(t)| < M_{x}$ for all $t \in I_{x}$. The collection of all such intervals $I_{x}$ forms an open cover of $[a, b]$ and by Heine Borel Theorem there are a finite number of these intervals, say, $I_{x_{1}}, I_{x_{2}}, \ldots I_{x_{n}}$ which cover $[a, b]$. Let $M = \max(M_{x_{1}}, M_{x_{2}}, \ldots, M_{x_{n}}$. Clearly every point $t$ of $[a, b]$ lies in some $I_{x_{j}}$ and hence $|f(t)| < M_{x_{j}}$ and thus $|f(t)| < M$ for all $t \in [a, b]$.
