properties of sets involving functions Suppose f is a function from X to Y and A, B are subsets of X, and suppose that S, T are
subsets of Y.
prove that $f(A\cap B) \subseteq f(A) \cap f(B)$.
my working:
let $x\in A\cap B$, then $f(x) \in f(A \cap B)$.
$x\in A$ and $x\in B$ so $f(x) \in f(A)$ and $f(x) \in f(B)$.
so $f(x) \in f(A) \cap f(B)$
Hence the forward direction is proved. However, somehow, the backward direction can also be proved.
let $x\in A$ and $x\in B$, so $f(x)\in f(A)\cap f(B)$
$x\in A\cap B$ so $f(x) \in f(A \cap B)$
How come the answer isn't an equal sign, and where did I go wrong, ie which step was invalid?

Consider the example of the $f(x) = \sin(x) $ for $A = [0, 2\pi]$ and $B=[-2\pi, 0 ] $.
$$A\cap B = \{0\} $$
$$f(A) = [-1,1] $$
$$f(B) = [-1,1] $$
$$f(A\cap B) = f(\{0\}) = \{0\}$$
$$f(A)\cap f(B) = [-1,1] $$
Of course $\{0\} \not= [-1,1]$.
