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Prove that a mapping f : S ! T is injective if and only if
f (A) intersection with f (B) = f (A intersection with B) for all A; B S.
P/s: sorry , I can't find where is the stupid symbol " intersection "
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Now assume f is injective. We show f(A) intersect f(B) is a subset of f(A intersect B): If y is in f(A) inter f(B), then y is in f(A) and y is in f(B). So y=f(x) for some x in A, and y = f(z) for some z in B. Since f is injective, x=z. Thus x is in A inter B, and f(x) is in f(A inter B)
Now assume f (A) intersection with f (B) = f (A intersection with B) for all A, B. We show f is injective: Suppose f(x)=f(z)=y. Let A={x}, B={y}. Then f(A inter B)=f(A) inter f(B) = {f(x), f(z)}={y}. So A inter B is not empty. Thus x = y. So f is injective.
Last edited by DrSteve (2010-12-13 23:54:00)
If you're going to be taking the SAT, check out my book:
http://thesatmathprep.com/SAT_Sales_Page.html
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