Hi, I am having trouble with these proofs; I don't know if I am doing these right. I'd appreciate some help. Thank you.(adsbygoogle = window.adsbygoogle || []).push({});

If X---> y is a map, then let B1, B2, B [tex]\subseteq[/tex] X.

i. f(B1 U B2) = f(B1) U f(B2)

To prove this I have:

f(B1 U B2)=f(B1) U f(B2)

Since B1 U B2 [tex]\subseteq[/tex] B1, we find that f(B1 U B2) [tex]\subseteq[/tex] f(B1) by this rule B1 [tex]\subseteq[/tex] B2 implies that f(B1) U f(B2).

Simarlarly, f(B1 U B2) [tex]\subseteq[/tex] f(B2). Hence f(B1 U B2) [tex]\subseteq[/tex] f(B1) U f(B2).

We have to assume that f is one to one, and y [tex]\in[/tex] f(B1) U f(B2) . So there are elements x1 [tex]\in[/tex] B1 and x2 [tex]\in[/tex] B2 with y=f(x1) and y=f(x2). Since f is one=to one, we conclude that x1=x2 [tex]\in[/tex] B1 U B2. Hence y [tex]\in[/tex] f(B1) U f(B2)

Assume equality holds for all choices of B1 and B2, and that f(x1)=f(x2). To show that x1=x2. Let B1={x1} and B2={x2}

f(B1) = {f(x1)}

= {f(x2)}

= f(B2)

Therefore,

f(B1 U B2) = f(B1) U f(B2) = {f(x1)} is not equal to the empty set

B1 U B2 is not equal to the empty set because x1=x2

ii.

B [tex]\subseteq[/tex] f^-1(f(B))

assume that f is one-to-one y [tex]\in[/tex] f^-1(f(B))

so there are elements x [tex]\in[/tex] B with y = f^-1(f(x)) implies that y [tex]\in[/tex] x

f is one to one, x [tex]\in[/tex] B, hence y [tex]\in[/tex] B.

iii.

f(X\B) [tex]\subseteq[/tex] Y\f(B) holds for all subsets B [tex]\subseteq[/tex] X if and only if f is one-to-one.

y [tex]\in[/tex] (Y\f(B) implies that y [tex]\in[/tex] Y and y [tex]\notin[/tex] f(B)

x [tex]\in[/tex] (X\f(B) implies that x [tex]\in[/tex] Y or f(x) and y [tex]\notin[/tex] f(B)

Therefore, x [tex]\in[/tex] f(x) and not in f(B) for all subsets B [tex]\subseteq[/tex] X. Thereore f(x) [tex]\subseteq[/tex] y

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# Proofs involving subsets

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