Disprove that "If $x$ is the supremum of $A$, then $f(x)$ is the supremum of $f(A)$"

Question

I'm still unexperienced with proving these kind of statements. So help will be appreciated.

Definition and properties of supremum :

If a non-empty set A $\subset \mathbb R$ is upper-bounded, then is $\xi \in \mathbb R$ the supremum if and only if the following properties hold:

(i) For all $x \in A$ holds that $x \le \xi$

(ii) For all $\epsilon \gt 0$ there exists an $x_\epsilon \in A$ met $\xi -\epsilon \lt x_\epsilon$.

If $x$ is the supremum of $A$, then $f(x)$ is the supremum of $f(A)$.

Is it sufficient to give the following function as an example to show the above statement is false? Because $A$ would than be $\arctan(x)$ and $f(x)= \tan(x)$ and $f(A)= \tan(\arctan(x))=x$

Answer 1

Yes, an example is enough. But I don't understand your example. What is the domain of $f$, for instance? What is $A$? It should be a set.

An example is $f\colon\mathbb{R}\longrightarrow\mathbb R$ defined by $f(x)=-x$ and $A=[0,1]$. Then $\sup A=1$, $f(\sup A)=-1$, and $\sup f(A)=0$.

Answer 2

$\arctan$ is not a set, it's a function, so it makes no sense to say that $A$ is $\arctan(x)$ since $A$ is required to be a non-empty subset. The $\sup$ is something a set has.

The $\tan$ function is not defined on the whole of $\mathbb{R}$. Yet, let us consider the set $U \subset \mathbb{R}$, which is the set on which $\tan$ is defined.

Then, $U$ is a set, and $\sup U = \infty$, since we can plug arbitrarily large values into the $\tan$ function. $\tan(U)$ is the set of values that the $\tan$-function attains. This is also a set, and $\sup \tan (U) = \infty$ since $\tan(x)$ can be an arbitratily large value.

If you have no restrictions on $f$, then counter-examples are abound. Define

$$f: [0,1] \to \{0, \alpha\} $$

for some $0 < \alpha \in \mathbb{R}$, and map $1 \mapsto 0$. Then $\sup ([0,1]) = 1$, and $\sup f([0,1]) = \alpha$, but $f(1) = 0 \neq \alpha$

Answer 3

To offer another simple counterexample, let $A = (-2, 2)$ and consider $f\colon A \to \mathbb{R}$ defined by $f(x) = 4 - x^{2}$. The supremum of $A$ is $2$. The graph of $f$ is below.

It's easy to see that the $f(A) = (0, 4)$ and that the supremum of $f(A)$ is $4 \neq 2$.