Bounded set and continuous function Let $f:\mathbb{R}^n \to \mathbb{R}^m$ continuous in $\mathbb{R}^n$. I want to proof that if $X \subset \mathbb{R}^n$ is a bounded set, then $f(X) \subset \mathbb{R}^m$ is also bounded.
I know that $\overline{X}$, the closure of X, would be compact, and then $f(\overline{X})$ should be compact too. So $f(X) \subset f(\overline{X})$ should clearly be bounded.
But I have to proof that proposition without using any compactness at all.
Any leads?
 A: Suppose that $f$ is unbounded. Then, for each $^k\in\Bbb N$, there is some $p_k\in X$ such that $\|f(p_k)\|\geqslant k$.
Now, let us write $p_k$ as $(a_{1k},a_{2k},\ldots,a_{nk})$. Since $X$ is bounded, the sequence $(a_{1k})_{k\in\Bbb N}$ has a convergent subsequence $(a_{1k_j})_{j\in\Bbb N}$; let $a_1$ be its limit. The sequence $(a_{2k_j})_{j\in\Bbb N}$ is also bounded; so, it has a convergent subsequence $(a_{2l_j})_{j\in\Bbb N}$; let $a_2$ be its limit. And you still have $\lim_{j\to\infty}a_{1l_j}=a_1$.
After having done this $n$ times, you get a subsequence $(p_{r_k})_{k\in\Bbb N}$ which converges to $a=(a_1,a_2,\ldots,a_n)$. But then, since $f$ is continuous, $\lim_{k\to\infty}f(p_{r_k})=f(a)$. This is impossible, since $\|f(p_{r_k})\|\geqslant r_k\geqslant k$ for each natural $k$ and therefore $\lim_{k\to\infty}\|f(p_{r_k})\|=\infty$.
A: Let $x \in \mathbb{R}^n$ and $f$ a continuous function on $\mathbb{R}^n$, suppose $f$ is unbounded, thus you have some sequence $x_n$ in $X$ for which the limit of $f(x_n)$ as n goes to infinity is $\infty$. Since f is continuous everywhere it follows that the closure of X must be unbounded. However it is easy to show that the closure of a bounded set is bounded in $\mathbb{R}^n$
