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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.

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this is indeed true see Andres answer –  Jr. May 4 '12 at 1:56

4 Answers 4

up vote 10 down vote accepted

Jr., I am writing a full answer, since the answers so far seem misleading. If $X$ is bounded, so is its closure $\bar X$, which is then a compact set. By continuity of $F$, $F(\bar X)$ is compact and therefore bounded. Since $F(X)\subseteq F(\bar X)$, then $F(X)$ is bounded as well.

As pointed out by others, the result is false in general without the assumption that $F$ is defined on all of ${\mathbb R}^m$.

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thank you for the explanation :) –  Jr. May 4 '12 at 1:59

Complete Revision of Earlier Answer: If $F$ is defined on all of $\Bbb R^m$, the statement is indeed true. If $X$ is bounded, then $\operatorname{cl}X$ is also bounded and therefore compact, and since $F$ is continuous, $F[\operatorname{cl}X]$ is a compact and hence bounded subset of $\Bbb R^n$. Since obviously $F[X]\subseteq F[\operatorname{cl}X]$, the result follows.

My original answer, based on a too-hasty reading of the question, noted that if $F$ is not required to be defined on all of $\Bbb R^m$, the result is false even for $m=n=1$, a simple counterexample being $X=(0,1)$ and $F(x)=1/x$.

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@Dylan: True, but I ought to fix it anyway! –  Brian M. Scott May 4 '12 at 1:50
    
But now $F$ isn't continuous! I think there is a difficulty after all. –  Dylan Moreland May 4 '12 at 1:54
    
@BrianM.Scott now this $F$ is not continuous... –  Jr. May 4 '12 at 1:56
    
@Dylan: And in fact I was being an idiot, because if $F$ is required to be defined and continuous on all of $\Bbb R^m$, the result is actually true. –  Brian M. Scott May 4 '12 at 2:05

Probably you meant that the domain is closed and bounded, in which case by the Heine-Borel theorem it $X$ is compact. Any continuous function on a compact domain attains a maximum and minimum, and hence is bounded. You can prove this using the Bolzano-Weierstrass Theorem.

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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.

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