Suppose that $f: X \rightarrow Y$ and that $x \in X $.
Definition 5.1.1 states that we let $X$ and $Y$ be sets. A function $f$ from $X$ to $Y$ written $f: X \rightarrow Y$ is a rule that pairs an element $x \in X$ with an element $y \in Y$, written $f(x) =y$ such that the following property holds.
$(( \forall x \in X)( \exists! y \in Y)[f(x) =y]$
Justify the statement: $x \in f^{-1}({f(x)}).$
$X$ belongs in the inverse of $f(x)$. The $f(x)$ is the image of elements in the domain. By Definition 5.1.8, if we let $f : X \rightarrow Y$ then the inverse of $f$ (or $f$ inverse denoted $f^{-1})$ is the pairing defined by the rule that if $f(x)=y$, then $f^{-1}(y)=x$.[...] all images of elements of the domain are in ${f(x) : x\in X}$
Is ${x}= \bar f^{-1} ({f(x)})$ true?
There is supposed to be {} on the x and f(x), but the latex wouldn't show it.
By definition 5.1.5, let $f: X \rightarrow Y$. The range of $f$ is the set ${y \in Y: (\exists x \in X), f(x)=y}. $
Recall that all images of elements of the domain are in ${f(x) : x\in X}$ . By definition 5.3.8, if we let $f: X \rightarrow Y$ for each set $B \in \mathcal P \left({Y}\right)$, then the function $\bar f^{-1}: \mathcal P \left({Y}\right) \rightarrow \mathcal P \left({X}\right)$ is defined by $\bar f^{-1} (B) = [x \in X: f(x) \in B]$.
I'm not even sure if I'm on the right track, but I have included some definitions that may be relevant to answering this question. For the second part, power sets are involved. So is it the power set {x} must equal or not equal to the power set inverse of the function on the right?!
\{
is needed for braces, since braces by themselves do grouping things. $\endgroup$