Homework Thread for Idiots

>>192719
It does seem logical indeed, at least for the => implication.

Undergrad math is far behind me and I'm gonna go to bed because it's late in my timezone, but the key to proving this equivalence lies very probably in the definition of bijectivity (or injectivity and surjectivity): I can distinctly remember that there is such a definition expressed in terms of element image/antecedent (f is bijective <=> forall y in Target set, there is a unique x in Source set such that y=f(x)) and another, equivalent one in terms of subset image/antecedent (but I can't remember it)

Perhaps find that particular definition in your textbook and try to see what you can get from it?

Good luck, oh and G \in P(A) is just a convoluted way of saying that G is a subset of A of course (I hate this P(A) notation)

>>192719

I just gave it a quick look, and the <= implication can be proved if u choose a G =A I think. For the other one maybe do what >>192729 suggests, I don't really know and would also need to remember the definitions to do it.

The function is bijective (one-to-one and onto or one-to-one correspondence) if every element of the codomain is mapped to by exactly one element of the domain. (That is, the function is both injective and surjective.) A bijective function is a bijection.

The function f: x--->y is bijective iff for all y element of Y, there is a unique x element of X such that f(x)=y.

>>192736
ok gave it another thought and the => can be solved easily if you use reductio ad absurdum (not sure if this si the correct english term).

Assume there is an f: A->B which is bijective but f(A\G) is not equal to B\f(G)

>>192747
I am not entirely sure what you're trying to state here. Are you saying that I should do a proof by contradiction to show that the opposite is true? If so, how would one do that?

>>192719
Could something be done knowing that the cardinalities of two sets of a bijection is identical. So removing G's elements from A would reduce the amount of elements... but I guess there is no implication that removing G elements from A would also remove an equal elements from B. Unless there is a way to prove it.

>>192736
A\G is substraction right?, the choossing G=empty set you solve the <=

>>192748
yes its proof by contradiction

>>192757
How so? If you do not mind me asking.

>>192760
since the statement says for all G in P(A) then you can choose G={ }

then f(A\G)= B\f(G) turns into f(A)=B which is the definition of bijectivity

>>192719
To prove the ⇐ part, prove injectivity and surjectivity independently. To prove injectivity, let a_1, a_2 be in A such that f(a_1) = f(a_2); show that a_1 = a_2 by considering G = { a_1, a_2 }. To prove surjectivity, consider G = {}.

To prove the ⇒ part, independently show that x in f(A\G) ⇒ x in B\f(G), and x in B\f(G) ⇒ x in f(A\G).

>>192768
Everything makes a lot of sense, but
>To prove the ⇒ part, independently show that x in f(A\G) ⇒ x in B\f(G), and x in B\f(G) ⇒ x in f(A\G).
How would one do this? I'm thinking of proving them using properties of bijectivity, but I am not entirely sure. In addition to that, could I possibly use subset properties?

>>192774
>I'm thinking of proving them using properties of bijectivity
Precisely.

For example, if x in B\f(G), then x in B and x not in f(G); and if x in B, then by surjectivity there is an y in A such that f(y) = x. Continue from here on.

OP here,

I have more or less solved what needed to be done. I had around 5 or 6 proofs in my head that were swirling around, and none of them seemed to work. Through reading what transpired here, I was able to fuse a few elements of each of them and narrow it down to 2 proofs, both of which are equally valid, I presume.

I want to thank everyone who ended up contributing to the thread, and I hope that the thread continues to live on for at least a little while to help other idiots in need of homework help.