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Given a set $A \subseteq \mathbb{R}$ and a continuous function $f$, does $A$ being bounded imply that $f(A)$ is bounded?

For $f(A)$ being non-bounded the condition $\exists s \in \mathbb{R} \forall a \in A: |f(a)| < s$ had to be broken. In a coninuous function that would mean that either $\limsup f(A) = \infty$ or $\liminf f(A) = -\infty$ or both, which is, as I think, not possible with a bounded domain since the function "ends" somewhere and thus a bigger or smaller value can respectively always be found in $\mathbb{R}$.

Am I right? And if I am, how can I express my thoughts in a mathematical way?

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    Do you mean *bounded*? – 2017-01-08
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    I am not a native speaker so I had to translate the term into english. The dictionary showed me "restricted". Is bounded correct? – 2017-01-08
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    If you mean that there are $u,v\in\mathbb{R}$ such that $u\le x\le v$ for all $x\in A$, then the right word is *bounded*. Is your native language Portuguese? – 2017-01-08
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    Is $f$ a function on $\mathbb{R}$, or just a function on $A$? $f$ is real-valued, I assume? – 2017-01-08
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    @egreg no I am not Portuguese. – 2017-01-08
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    @Hurkyl yes $f$ is a function on R – 2017-01-08

3 Answers 3

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If $f$ is assumed to be a continuous function on $\mathbb{R}$, then yes: if $A$ is bounded then $f(A)$ is bounded.

This is because a nonempty $A\subseteq\mathbb{R}$ is bounded if and only if its closure is bounded (that is, compact) and the image of a compact subset through a continuous function is again compact, hence closed and bounded. So, if $A$ is bounded, we have $f(A)\subseteq f(\overline{A})$ and so $f(A)$ is bounded.

If the function is defined on a subset of $\mathbb{R}$, then this is no more true: consider $f(x)=\tan x$ on $(-\pi/2,\pi/2)$.

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    These are important points. The domain and codomain of $f$ *need* to be specified in order for this question to be answerable (which you've covered here by looking at both cases). – 2017-01-08
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consider the function $f:(0,1)\rightarrow \mathbb R$ given by $f(x)=\frac{1}{x}$.

In fact, a set $A\subseteq \mathbb R$ satisfies that all the continuous functions $f(A)\rightarrow \mathbb R$ are bounded if and only if $A$ is compact.

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You can show first that if $f$ is a continuous real function then for each $r>0$ the set $\{f(x): |x|\leq r\}$ is bounded. So if $A\subset \mathbb R$ and $A$ is bounded then there exists $r>0$ with $A\subset [-r,r].$ Therefore $\{f(x):x\in A\}$ is bounded because it is a subset of the bounded set $\{f(x):|x|\leq r\}.$

To prove that $\{f(x): |x|0:$

Suppose, by contradiction, that it is not.

Let $a_1=-r$ and $b_1=r$ and $I_1=[a_1,b_1=[-r,r].$ Recursively, when $I_n=[a_n,b_n]\;$ let $I_{n+1}=[a_n,(a_n+b_n)/2]\;$ if $\;\exists x\in [a_n,(a_n+b_n)/2]\;(|f(x)|>n),$ and if not, then let $I_{n+1}=[(a_n+b_n)/2,b_n].$

So $\exists x\in I_n\;(|f(x)|>n).$

By induction on $n$ we have $|a_n-a_{n+1}|\leq (2r)2^{-n}$ and $|a_n-b_n|=(2r)2^{1-n}.$

So there exists $ L\in [-r,r]$ with $L=\lim_{n\to \infty}a_n=\lim_{n\to \infty}b_n.$

Now for any $\epsilon >0 $ there exists $n\in \mathbb N$ such that $n>|f(L)|+1$ and $\max (|L-a_n)|,|L-b_n|)<\epsilon /2,$ so that $[a_n,b_n]\subset (-\epsilon +L,\epsilon+L).$

But $\exists x\in [a_n,b_n]\;(\;|f(x)|>n>|f(L)|+1\;).$ This gives the desired contradiction: Any neighborhood of $L$ contains a point $x$ with $|f(x)-f(L)|>1,$ implying that $f$ cannot be continuous at $L.$