$A$ is bounded$\Rightarrow f(A)$ is bounded. 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?
 A: 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 $(-\pi/2,\pi/2)$.
A: 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.
A: 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 $\{f(x): |x|<r\}$ is bounded when $f$ is continuous and $r>0:$
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 $L.$ 
