I'm having difficulty completing a proof of a statement (my attempt is written after the statement):

Given a monotonically increasing function $f:\mathbb R \rightarrow \mathbb R$, and a bounded set $A\subset \mathbb R$, I am trying to prove that


where the $f(A)$ is the set defined by $f(A)=\{f(a) : \,a\in A\}$.

I managed to prove that $\inf(f(A)) \geq f(\inf(A))$:

Let $a\in A$. Then by definition of $\inf$:

$$a\geq \inf(A)$$

Because $f$ is monotonically increasing, for any $a,b\in \mathbb R$ such that $a\geq b$, $f(a) \geq f(b)$.


$$\forall a\in A\quad f(a)\geq f(\inf(A))$$

Which means that

$$\inf(f(A))\geq f(\inf(A))$$

I am struggling with showing the reverse inequality (or show a contradiction of the strong inequality). I tried using the definition of $\inf$ with $\epsilon$ but I didn't see anything that might help with the proof.

So, how might I prove the rest of the statement? (And is my proof correct, up till this point?)

I have searched around and haven't found something about the image of a set (mapped by a monotone function).


It seems that the statement isn't true (an answer gives a counterexample). So, I suppose if $f$ is continuous then it is true, but I haven't checked.

  • $\begingroup$ Maybe $A$ should be also closed? $\endgroup$ Jan 27, 2018 at 10:08
  • $\begingroup$ @Taroccoesbrocco I think if A is closed it still won't be enough. After seeing Jose's answer, $f$ probably needs to be continuous for that to be true. $\endgroup$
    – ItamarG3
    Jan 27, 2018 at 10:09

1 Answer 1


You can't prove it, because it is false. Take$$f(x)=\begin{cases}x&\text{ if }x>0\\x-1&\text{ otherwise}\end{cases}$$ and $A=(0,1)$. Then$$\inf f(A)=0\text{ and }f(\inf A)=-1.$$

  • $\begingroup$ $A$ is bounded. the example you gave isn't bounded. $\endgroup$
    – ItamarG3
    Jan 27, 2018 at 10:05
  • $\begingroup$ @ItamarG3 That changes nothing. I've edited my answer. $\endgroup$ Jan 27, 2018 at 10:06
  • $\begingroup$ I see. One moment... $\endgroup$
    – ItamarG3
    Jan 27, 2018 at 10:06
  • $\begingroup$ Ok. Thanks. I keep forgetting it doesn't have to be continuous. $\endgroup$
    – ItamarG3
    Jan 27, 2018 at 10:08
  • $\begingroup$ the accept seems to have failed when I tried to mark it... problem gone now. o.o $\endgroup$
    – ItamarG3
    Jan 27, 2018 at 11:31

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