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...
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...
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...
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?