Recent content by Uncanny

Here’s an update with things written out: Definition: Let ##G## be a graph. ##G## is a functional graph if and only if ##(x_1,y_1) \in G## and ##(x_1,y_2) \in G## implies ##y_1=y_2##. Problem statement, as written: Let ##G## be a functional graph. Prove that ##G## is injective if and only if...

Ah, I think I understand now. If A is empty and B is non-empty, both directions of the biconditional result in vacuous truth. If we choose A empty, but not B, then f is not surjective, making the antecedent false. Analogous argument going in the other direction, per your reasoning in post #9...

😅 thanks, I fixed that. Thanks for your reply. ##J## and ##H## are actually only given to be arbitrary graphs. I attached a photo of the page the problem appears on (I’m working out of Charles Pinter’s “Book of Set Theory”) so you can verify this. I have a feeling the premise “G must be a...

Thanks! You hit the nail on the head with the case in 17b) in which B is non-empty, but A is. This is the case I was concerned with. I agree with everything you’ve written, I suppose my “real” question, then, is: shouldn’t the problem statement avoid or warn against this outcome by including...

Ah shoot.. sorry guys. Figuring out the formatting

Definition: Let ##G## be a graph. ##G## is a functional graph if and only if ##(x_1,y_1) \in G## and ##(x_1,y_2) \in G## implies ##y_1=y_2##. Problem statement, as written: Let ##G## be a functional graph. Prove that ##G## is injective if and only if for arbitrary graphs ##J## and ##H##, ##G...

17b) also requires A to be non-empty though, no? Otherwise, the definition of surjection can’t be satisfied.

it sounds like you talked yourself into a hole you’ve realized you can’t dig yourself out of

Let’s try this instead: I challenge anyone to a proof of the surjectivity of f without resorting to placing any restrictions on C.

It just means g = h = {(b1,c), (b2,c)} in that case. I’m showing that if C is allowed to be a singleton, a counterexample can be constructed.

Tell you what, find me a proof where C may be a singleton. 👍

I’m working in the “if” direction, so I’m trying to “prove” f is surjective. I assume for any g,h etc... If I suppose it’s true for any g,h that etc.. then I can pick whichever I like, and the property will hold. Again: I assume it’s true for all, so it’s true for any particular one (or two). I...

Consider C = {c}, B = {b2, b1}. Let g = h = B x C. f need not be surjective, even though g o f = h o f implies g = h.

Actually, let’s more carefully consider the set C. If C is a singleton, a contradiction may be reached. Thus, perhaps there is, indeed, a mistake present: C must be restricted be contain at least two (distinct) elements. It does seem strange that this statement is omitted, since the earlier...

Alright, gotcha. That makes sense now- so that’s the more practical reason the restriction is necessary: when, without it, the intended result(s) fails to hold?