TITLE: Continuous functions are bounded
QUESTION [10 upvotes]: How one demonstrates this claim:
Let $F:\mathbb{R}^m \rightarrow \mathbb{R}^n$ be a continuous function, if $X \subset \mathbb{R}^m$ is bounded then $F(X)$ is bounded.

REPLY [3 votes]: Just in case you want a solution that doesn't use Heine-Borel etc, here's a somewhat more elementary proof.  Suppose $F(X)$ is not bounded.   Since $X$ is bounded, $X$ is contained in the union of some finite collection of $m$-cubes $\{x: a_i \le x_i \le a_i+1\ \text{for}\ i=1\ldots m\}$  (say with the $a_i$ integers).  Therefore $F$ would have to be unbounded on at least one of these $m$-cubes, say $C_0$.  Now $C_0$ is the union of $2^m$ $m$-cubes of side $1/2$, so $F$ is also unbounded on at least one of those, say $C_1$.  Inductively, we get a nested sequence of $m$-cubes $C_k$ of side $2^{-k}$ with each $F(C_k)$ unbounded.
Picking a point in each $C_k$ gives us a Cauchy sequence, which converges to some point $p$ of ${\mathbb R}^m$.  Since $F$ is continuous at $p$, there is some $\delta > 0$ such that
for every $x$ with $\text{dist}(x,p) < \delta$, $|F(x) - F(p)| < 1$.  In particular, 
$F$ is bounded on the open ball $B$ of radius $\delta$ centred at $p$.  But for $k$ sufficiently large, $C_k$ is contained in $B$, and we get a contradiction since $F(C_k)$ is supposed to be unbounded.
You can even get away without appealing to completeness of ${\mathbb R}^m$ if you accept the fact that real numbers can be specified by infinite strings of digits $a.d_1 d_2 d_3 \ldots$: instead of splitting $C_k$ into $2^m$ sub-cubes of side $2^{-k}$, split it into $10^m$ sub-cubes. Each $C_k$ is of the form $\{x: a_i \le x_i \le a_i + 10^{-k} \ \text{for}\ i=1\ldots k\}$ where $a_i$ is a number with $k$ digits after the decimal point (for simplicity I'll assume $a_i \ge 0$), and the $a_i$ for $C_{k+1}$ is obtained by appending another decimal digit 0 to 9 to the $a_i$ for $C_k$.  The $i$'th coordinate of the point $p$ then is the real number that agrees with the $a_i$ of each $C_k$ up to the $k$'th digit after the decimal point.