Question

Need some help on a couple proofs with functions

Answer 1

Should be along the lines of the pigeonhole principle I assume

Answer 2

check this
http://math.stackexchange.com/questions/170725/do-we-have-always-fa-cap-b-fa-cap-fb

Answer 3

I guess that makes sense for a) , not sure how you can create a counter example for b) however

Answer 4

Any ideas?

Answer 5

for a counterexample, look at marlu's answer in that link

Answer 6

Consider the case where f is constant and A and B are disjoint.

Answer 7

f(x) = 1
A = {2, 3}
B = {4, 5}

Answer 8

\(f(A)\cap f(B) = \{1\}\)

\(f(A\cap B) = ?\)

Answer 9

there are no elements that A intersect B has so f(A intersect B) should be empty?

Answer 10

Yes

\(f(A)\cap f(B) = \{1\}\)

\(f(A\cap B) = f(\emptyset) = \emptyset\)

Answer 11

they are not same.

recall that \(\emptyset\) is a subset of everyset, so we just have \(f(A\cap B)\subseteq f(A)\cap f(B)\)

Answer 12

Alrighty that makes sense, thanks Ganeshie

Answer 13

np