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# some pure maths set theory and functions problems watch

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1. 1. let f:X-->Y be a function, and let A,B be subsets of X and C,D be subsets of Y. give proofs or counterexamples of
a) f(A intersection B)=f(A) intersection f(B)
b) inverse f(C intersection D)= inverse f(C) intersection inverse f(D)

2. prove that the set of finite subsets of the natural numbers is countable

3. let A be a finite set with n elements. prove that the power set P(A) has 2^n elements.

4. prove by strong induction that for every integer n greater than or equal to 2, either n or n-1 is a sum of distinct primes.

help would be appreciated
xx
2. what college are you at? I'm (supposed to be) doing that problem sheet.
3. hertford. the problem sheet is TERRIBLE i cant do most of it. pfft. what college you at?
4. lincoln. i didn't think i could do any of it, but its alright once i tried it for a bit. still cant do 4b
5. Doing them one at a time

1a. False check

1b. by 1a
Hence

not 100% certain on this proof

2. countable union of countable sets

3. Each of the n elements is either a member of the set or isn't, so 2^n possible sets in power set.

4. Not sure, seems easy at first but proof evades me, it may be one of those questions that seems so obvious but is actually very hard, an analogous task to the infamous question 14 on the Cambridge Numbers and Sets example sheets, either that or I just can't spot it.
6. (Original post by apd35)
Doing them one at a time

1a. Suppose then and hence
no, 1a is false

counterexample:

let f(x) = x^2

A={1}
B={-1}

f(AnB)={empty}

f(A)nf(B)={1}
7. Fair enough, it's late and I should be in bed, in that case I'm more certain on my 1b proof, I felt it should be true one way and not the other.

Mine assumes that there exists
8. (Original post by emma4888)
1. let f:X-->Y be a function, and let A,B be subsets of X and C,D be subsets of Y. give proofs or counterexamples of
a) f(A intersection B)=f(A) intersection f(B)
b) inverse f(C intersection D)= inverse f(C) intersection inverse f(D)
APD: for part a) you need f to be injective, Did not see your proof, so cant say where it is needed. If you still have your proof you should be able to see where this condition is needed.
b)
let t be in f^(-1){CnD}
<=> f(t) is in CnD
<=> f(t) is in C and f(t) is in D
<=> t is in f^(-1) {C} n f^(-1) {D}
following arrows from top to bottoms gives
f^(-1){CnD} is contained in f^(-1) {C} n f^(-1) {D}

following arrows from bottom to top gives
f^(-1){CnD} contains f^(-1) {C} n f^(-1) {D}

hence
f^(-1){CnD}=f^(-1) {C} n f^(-1) {D}

4) true for 2 obviously.
assume for all integers 2,3,,,,,k
consider k+1
if K+1 is prime or K is prime we are done. assume not
let P be the biggest prime st P<k+1
then by induction since
K+1-P>=2
we can write
K+1-P as sum of distinct primes.
K+1-P=p1+p2+p3+...+pk
ie
K+1=p1+p2+.....pk+P

or (K+1-P)-1 can be written as sum of distinct primes
we can write
K-P as sum of distinct primes.
K-P=p1+p2+p3+...+pk
ie
K=p1+p2+.....pk+P

and these are distinct by choice of P

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