Prove that $X \subset Y \implies f^{-1}(X) \subset f^{-1}(Y)$ Let E and F be two sets and $f: E \to F $ be a function, and $X, Y \subset F$. 
Prove that $X \subset Y \implies f^{-1}(X) \subset f^{-1}(Y)$
My answer: 
Let $y \in X$, then $f^{-1}(y) \in f^{-1}(X)$.
Since $X \subset Y$, then $y \in Y$.
If $y \in Y$, then $f^{-1}(y) \in f^{-1}(Y)$.
So, $f^{-1}(y) \in f^{-1}(X)$ and $f^{-1}(y) \in f^{-1}(Y)$, then $f^{-1}(X) \subset f^{-1}(Y)$.
Is this proof ok?
 A: One thing that come to mind is that you have to think carefully about using $\in$ and $\subset$. Unless your function is bijective it is a very real possibility that $f^{-1}(y)$ is not an element, but a set. In this event the statement $f^{-1}(y) \in Y$ is not true.
A: The proof looks pretty good! There are some things that I can say though. 
For starters, we want to show the implication $x\in f^{-1}(X)\implies x\in f^{-1}(Y)$. Because of this, we don't really need to look for $y\in Y$. You made this claim, and then the corresponding claims about the preimages with a subtle implication to if $y\in f^{-1}(X)\cap f^{-1}(Y)$ then $f^{-1}(X)\subset f^{-1}(Y)$. But this actually doesn't follow. See if you can revamp it.
If you'd like, I've rewritten it below but, I would only look at this after trying again first.
Suppose $X\subset Y$ (both subsets of the codomain, $F$), and take an element $x\in f^{-1}(X)$. Then $f(x)\in X$ by the definition of preimage. Since $X\subset Y$, we have $f(x)\in Y$. Therefore, if $x\in f^{-1}(X)$ then $x\in f^{-1}(Y)$ showing $f^{-1}(X)\subseteq f^{-1}(Y)$.
A: These are the standard definitions:
Let $X, Y$ be any sets whatsover. 
$$ f^{-1}(X) = \{ e \in E \ \mid \ f(e) \in X \}$$
$$ f^{-1}(Y) = \{ e \in E \ \mid \ f(e) \in Y \}$$
Your requirement is to prove that the former is a subset of the latter, given $X \subseteq Y$. So we prove the implication 
$$ a \in  f^{-1}(X) \implies a \in f^{-1}(Y)   $$
So to this end suppose $a \in  f^{-1}(X)$. The qualification for this is that $ a \in E $ and $ f(a) \in X $. But since, $X \subseteq Y$ any element in $X$ is also an element of $Y$. Hence $ f(a) \in Y$. 
So we can extract $ a \in E $ and $f(a) \in Y$ which puts $a \in f^{-1}(Y)$
$\mathscr{Q.E.D.}$
A: $$\begin{align}
x\in f^{-1}(X) & \implies f(x)\in X\\
& \implies f(x)\in Y\\
& \implies x\in f^{-1}(Y)
\end{align}$$
A: A={x in E : f(x) is in X} = f inverse of a set X
B={x in E : f(x) is in Y} = f inverse of a set Y
if X is subset of Y:
f(x) in X means f(x) in Y 
then A is subset of B
