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Let $f: \mathbb{R}^n \to \mathbb{R}^m$ be a continuous function, and $S \subset \mathbb{R}^n$ a bounded set. Now I want to show that $f(S)$ is bounded.

My thoughts: Since $S$ is bounded it's contained in an open ball $B(a,r)$ for some $r>0$. Then I want to show that $f(B(a,r))$ is a bounded set. I know from the continuity of $f$ that given $\varepsilon >0$ there is $\delta >0$ such that $f(B(a,\delta)) \subset B(f(a),\varepsilon)$ for any $a \in \mathbb{R^n}$. So can I just say I could pick the right $\varepsilon$ so that $\delta = r$? But I feel something is amiss here...

carmichael561
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meiji163
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2 Answers2

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I think compactness is needed here. Find some $M>0$ such that $S\subseteq K:=\{|x|\leq M\}$, $K$ is compact, and so is the image $f(K)$, hence it is closed and bounded, in particular, it is bounded. Now $f(S)\subseteq f(K)$, so $f(S)$ is bounded as well.

user284331
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If you add continuous functions $\color{magenta}{\text{on } \Bbb R^n}$ to your title, then it will be correct.

Continuous functions do not preserve boundedness. For instance, $\tan:(-\pi/2,\pi/2)\twoheadrightarrow \Bbb R.$

They do preserve compactness though. The proof is an easy exercise.

calc ll
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