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 itâ€™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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