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

...
Oct 17th, 2017
...
Oct 17th, 2017
Oct 18th, 2017
check_circle