If x is a subset of Y then, is inverse of x also a subset of Y?

Let $f : X \rightarrow Y$ be a function. If B is subset of y then its inverse image $f^-B$ is the subset of x define by $f^-B = {x : f(x) \in B}$

Now prove the following.

$i. B_1 \subset B_2 \Rightarrow f^-B_1 \subset f^- B_1$

$ii. f^-\cup_i B_i = \cup_i f^-B_i$

$iii. f^-\cap_i B_i \subset \cap_i f^-B_i$

Solution:

Let $F(x) : x \rightarrow y$ be a function. If B is the subset of Y, then it's inverse maps

$f^-x$ is the subset of x define by $f^-B = { f(x) \in B}$

i.

$let\ x \in f^-B_1\ then\ f(x) \in B_1$

$since\ B_1 \subset B_1\$ $we\ can\ write,\ f(x) \in f^-(B_2)$

ii.

$Suppose\ , x \in f^-\cup_i(B_i)$, $f(X) \in \cup_i(B_i)$

$So\ f(x) \in B_i for\ some\ i$ this implies that, $x \in f^-B_i$

$So\ x \in \cap_i f^-(B_i)$.

Conversely

$Suppose\ x \in \cap_i f^-(B_i)$. $Then\ x \in f^-(B_i)\ for\ some\ i$

$So\ f(x) \in B_i \subset (\cap_iB_i)$ therefore $f(x) \in (\cap_i B_i)$ and hence $x \in f^-(\cap_iB_i)$