Time remaining:
Introduction to Math Proof

label Mathematics
account_circle Unassigned
schedule 0 Hours
account_balance_wallet $5

Question No. 1: Suppose that S is a set. Prove that the subset relation ⊆ is a partial order on P(S), the power set of S.

Question No. 2: Suppose f: X→Y and let A⊆X and B⊆X. Prove:

f[A\B]⊇f[A]\f[B].

Give an example where equality fails.

Oct 17th, 2017

1. P(S) is the set of all subsets of S. We ought prove that for any A, B, C from P(S)
a) A in A. It is obviois: for any xϵA true that xϵA;
b) A in B and B In A => A = B. Really, for any xϵA we see xϵB because A in B. Also for any yϵB we see yϵA, so A = B;
c) for A in B and B in C => A in C. Again the proof looks like a simple speaking: for any xϵA we know xϵB (because A in B) and further xϵC as B in C.

2. Once again we rely on definitions only.
The left side: y ϵ f(A\B) means there exists aϵ(A\B) such that y = f(a) AND aϵ(A\B).
In turn, aϵ(A\B) means aϵA AND a not ϵ B.

The right side: z ϵ f(A)\f(B) means that zϵf(A) AND z not ϵ f(B).
this turns into: there exists aϵA such that z = f(a) AND for any bϵB z not= f(b).

So, let z ϵ f(A)\f(B), we ought prove that z ϵ f(A\B).
But "there exists aϵA such that z = f(a) AND for any bϵB z not= f(b)"
implies z=f(a) AND aϵA AND a not ϵ B (a not in B because for any a in B z not= f(a)).
So the proof is complete.

Example: consider f(x) = x^2, f from R to R.
Let A=[-1; 1] B= [0; 1] (now prove)


Apr 27th, 2015

now prove that f(A\B) greater than f(A)\f(B):

f(A) = [0; 1], f(B) = [0; 1], so f(A)\f(B) is empty.
But A\B = [-1; 0) and f(A\B) = (0; 1] -- not empty.

That's all.

Apr 27th, 2015

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Oct 17th, 2017
...
Oct 17th, 2017
Oct 18th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer