Infimium between two sets with a bijection

Question

People, I need some help about a thing: I am studying analysis and, between my thoughts about supremum and infimum, I started to ask myself the following:

Let A, B $\neq \emptyset$ and $f:A\rightarrow B$ be a bijection. Assume that there A and B are bounded below. Show that $f(\inf(A))=\inf(B)$.

Is that sentence true? Why? I did not found a way to prove it, since I can not assume that $f$ takes the infimium from one set to the another, once $\inf(A) \notin A$, in general.

Answer 1

I am interpreting "inferiorly limited" as "bounded below" since I haven't seen that terminology before; and this does not seem true at all.

Suppose $A = \{1,2,3\}, B=\{4,5,6\}$, then $\text{inf}(A) = min(A)= 1$, $\text{inf}(B)=min(B)=4$.

And we have the bijection $f: A\rightarrow B, f(x)=x+3$.

Giving us a contradiction.

Answer 2

Here's a contradiction: $A = B = [0, 1]$, and $f(x) = 1 - x$. Then $f(\inf A) = f(0) = 1 \neq \inf(B)$.