**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)$

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