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$f\colon\mathbb{R}^n\rightarrow \mathbb{R}^m$ is continuous; $A\subset \mathbb{R}^n$. Is bounded is $f(A)$ bounded?

I know that if the function was only from $A$ to $\mathbb{R}^m$ then it is false. However, since we have the condition that $f$ is continuous all over $\mathbb{R}^n$. I'm pretty sure $f(A)$ is supposed to be bounded.

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    Hint: $A \subseteq \overline{A}$ and what do you know about $f(\overline{A})$? – 2017-02-27

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If $f(A)$ is not bounded, then there exists a sequence $\{x_n\}\subset A$, such that $|f(x_n)|\to\infty$. But, as $\{x_n\}$ is bounded, it possesses a convergent subsequence $x_{k_n}\to x$. Since $f$ is continuous, $f(x_{k_n})\to f(x)$, which contradicts the fact that $|f(x_n)|\to\infty$.

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    Perhaps this comment is unnecessary, but it seems worthwhile to point out that $x$ may not lie in $A$. This is okay, since $f$ is assumed to be continuous on all of $\mathbb{R}^n$. I only mention this because one could easily be fooled into thinking that this argument works for a function which is only guaranteed to be continuous on $A$, which it does not. Nice answer! – 2017-02-28
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To put the other answer a bit differently: We have that $A \subseteq \overline{A}$, and since $A$ is bounded, $\overline{A}$ is compact. It follows that $f(\overline{A})$ is compact and hence bounded. Since we have that $f(A) \subseteq f(\overline{A})$, the result follows.

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Yes, this is true.

Since $f$ is defined on $\mathbb R^n$, and $A\subset \mathbb R^n$, let's take $F$ a close subset containing $A$ (you can take $F=\overline A$ the closure of $A$).

Then since $f$ is continuous on $F$, $f(F)$ is bounded because $F$ is closed.

So $f$ is bounded on $F$ so $f$ is bounded on $A$ because $A\subset F$.

So $f(A)$ is bounded.