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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