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Borg7

Math Problem

If f : X → Y is injective and A and B are both subsets of X, then f(A ∩ B) = f(A) ∩ f(B).

Does anybody know how I can prove the above? I've been searching for the solution for hours now, but I can't come up with it.
Brokerstar

If anyone knows the answer to that but can't figure out how to win at poker then I'm just a fish on a heater!!   Rolling Eyes
U Cook Socks

It's times like these, I realise how thick I really am. Cheers Borg.
The Angler

Assume f(A /\ B) = f(A) /\ f(B) and distinct x,y with f(x) = f(y).

Then nulset = f(nulset) = f({x} /\ {y}) = f({x}) /\ f({y}) = {f(x)}.
Take it from there to show f injective

Asume f injection.  You know f(A /\ B) subset f(A) /\ f(B).
If not equal, then some y in f(A) /\ f(B), with y not in f(A /\ B).
Thus some a in A, b in B with y = f(a) = f(b)
   and for all x in A /\ B, y /= f(x).
Hence a /= b and f isn't injection.

Conditions equivalent surjective f
   for all x in Y, f^-1(x) nonnul
   for all nonnul A subset Y, f^-1(A) nonnul
   for all V subset Y, ff^-1(V) = V
   for all U subset X, Y\f(U) subset f(X\U)

Conditions equivalent injective f
   for all x in Y, f^-1(x) empty or singleton
   for all U subset X, f^-1f(U) = U
   for all U,V subset X, f(U /\ V) = f(U) /\ f(V)
   for all U subset X, f(X)\f(U) = f(X\U)

Condition equivalent bijective f
   for all x in Y, f^-1(x) singleton

----

f surjection iff
    iff
f injection iff
   x in f'f(U) iff f(x) in f(U) iff some u in U with f(x) = f(u) iff x in U
   f(U/\V) = f(f'f(U) /\ f'f(V)) = ff'(f(U) /\ f(V)) = f(U) /\ f(V)
   nulset = f(U /\ X\U) = f(U) /\ f(X\U)
       f(X\U) subset f(X)\f(U) subset f(X\U)
   if f(x) = f(y):  f(y) in f({x}) = f(X\(X\x)) = f(X)\f(X\x)
       f(y) not in f(X\x);  y not in X\x;  y in {x};  x = y






from http://www.mathkb.com/Uwe/Forum.a...n-B-f-A-n-f-B-iff-f-is-one-to-one
Brokerstar

hieroglyphics

I never understood them.
Borg7

Haha @ broker and blazing

Thanks Angler, I got it now.
The Angler

Happy to help.
BetMagicMoney

i agree with the proof Very Happy
kolonel

The Angler wrote:
Assume f(A /\ B) = f(A) /\ f(B) and distinct x,y with f(x) = f(y).

Then nulset = f(nulset) = f({x} /\ {y}) = f({x}) /\ f({y}) = {f(x)}.
Take it from there to show f injective

Asume f injection.  You know f(A /\ B) subset f(A) /\ f(B).
If not equal, then some y in f(A) /\ f(B), with y not in f(A /\ B).
Thus some a in A, b in B with y = f(a) = f(b)
   and for all x in A /\ B, y /= f(x).
Hence a /= b and f isn't injection.

Conditions equivalent surjective f
   for all x in Y, f^-1(x) nonnul
   for all nonnul A subset Y, f^-1(A) nonnul
   for all V subset Y, ff^-1(V) = V
   for all U subset X, Y\f(U) subset f(X\U)

Conditions equivalent injective f
   for all x in Y, f^-1(x) empty or singleton
   for all U subset X, f^-1f(U) = U
   for all U,V subset X, f(U /\ V) = f(U) /\ f(V)
   for all U subset X, f(X)\f(U) = f(X\U)

Condition equivalent bijective f
   for all x in Y, f^-1(x) singleton

----

f surjection iff
    iff
f injection iff
   x in f'f(U) iff f(x) in f(U) iff some u in U with f(x) = f(u) iff x in U
   f(U/\V) = f(f'f(U) /\ f'f(V)) = ff'(f(U) /\ f(V)) = f(U) /\ f(V)
   nulset = f(U /\ X\U) = f(U) /\ f(X\U)
       f(X\U) subset f(X)\f(U) subset f(X\U)
   if f(x) = f(y):  f(y) in f({x}) = f(X\(X\x)) = f(X)\f(X\x)
       f(y) not in f(X\x);  y not in X\x;  y in {x};  x = y






from http://www.mathkb.com/Uwe/Forum.a...n-B-f-A-n-f-B-iff-f-is-one-to-one


forced

lmao!
sandman369

I would've been able to answer this earlier this year when I was in discrete math class... now it's all flown out of my squishy head blob thingy
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