# $f : \mathcal {P}(X) \rightarrow \mathcal{P}(Y)$ is monotonically increasing [closed]

Let $$X$$ and $$Y$$ be two sets such that $$f \in Y^X$$ then follows that those from $$f$$ induced functions $$f : \mathcal {P}(X) \rightarrow \mathcal{P}(Y)$$ and $$f^{-1} : \mathcal{P}(Y) \rightarrow \mathcal {P}(X)$$ are monotonically increasing.

My questions:

1) are the sets $$X$$ and $$Y$$ by nature ordered?

2) If so how does this order reflect on the induced functions from $$f$$ on the powersets? If not, how can an order on the induced functions $$f$$ even arise?

A nicely constructed example is welcome.

No, the sets are not ordered by nature. The power set of any set is naturally a poset with the inclusion relation; this doesn't reflect anything about the underlying sets. This (partial) order is defined as $$A \leq B$$ if and only if $$A \subseteq B$$. Note that not every pair of sets is comparable, for example $$\{1,2\}$$ is neither less than nor greater than $$\{2,3\}$$.
Any function $$f$$ from $$X$$ to $$Y$$ induces monotonic functions $$f$$ (I don't love this notation, since this is a different function from $$f$$) from $$\mathcal{P}(X) \rightarrow \mathcal{P}(Y)$$ and $$f^{-1}$$ from $$\mathcal{P}(Y) \rightarrow \mathcal{P}(X)$$. "Monotonic" in this case means that:
For any subsets of $$X$$, say $$X' \subseteq X''$$, $$f(X') \subseteq f(X'')$$
For any subsets of $$Y$$, say $$Y' \subseteq Y''$$, $$f^{-1}(Y') \subseteq f^{-1}(Y'')$$.
Here is a small example: Let $$X$$ and $$Y$$ both be the set of real numbers $$\mathbb{R}$$. Then the two intervals $$[1,2]$$ and $$[1,3]$$ are both subsets of the reals, making them elements of $$\mathcal{P}(X)$$ and of $$\mathcal{P}(Y)$$, and $$[1,2] \leq [1,3]$$ in the order on the powerset, because one is a subset of the other. Let $$f$$ be the function $$f(x) = x^2$$. Then $$f([1,2]) = [1,4]$$ and $$f([1,3]) = [1,9]$$. As expected, $$[1,4] \subseteq [1,9]$$, meaning that the order is preserved.
In the opposite direction, we see that $$f^{-1}([1,2]) = [-\sqrt{2},-1] \cup [1, \sqrt 2]$$, and $$f^{-1}([1,3]) = [-\sqrt{3}, -1] \cup [1, \sqrt{3}]$$. Again, the first set is a subset of the second, so the order was preserved.