Function equality

[Elwin.Martin]

1. The problem statement, all variables and given/known data
Two functions f and g are equal iff
(a) f and g have the same domain, and
(b) f(x) = g(x) for every x in the domain of f.

Not actually hw, but I wanted to prove or at least see the elementary proof of this theorem.
2. Relevant equations
just the formal function definition:
A function f is a set of ordered pairs (x,y) [such that] no two of which have the same first member.

y = f(x), customarily is used over (x,y) is an element of f
3. The attempt at a solution
Alright, I just want to know if I'm over-simplifying here...but,

Since f is a set of ordered pairs and g is a set of ordered pairs, for f = g we must have that every ordered pair in f is an ordered pair in g and every ordered pair in g is an ordered pair in f (subsets of one another). For f and g to be equal then, they must contain the same first members and thus the same domain.

And then then something along the same lines for the second part.

Is this too cheap or incorrect somewhere? I'm kind of awful at rigor so any help would be greatly appreciated.

[SammyS]

1. The problem statement, all variables and given/known data
Two functions f and g are equal iff
(a) f and g have the same domain, and
(b) f(x) = g(x) for every x in the domain of f.

Not actually hw, but I wanted to prove or at least see the elementary proof of this theorem.
2. Relevant equations
just the formal function definition:
A function f is a set of ordered pairs (x,y) [such that] no two of which have the same first member.

y = f(x), customarily is used over (x,y) is an element of f
3. The attempt at a solution
Alright, I just want to know if I'm over-simplifying here...but,

Since f is a set of ordered pairs and g is a set of ordered pairs, for f = g we must have that every ordered pair in f is an ordered pair in g and every ordered pair in g is an ordered pair in f (subsets of one another). For f and g to be equal then, they must contain the same first members and thus the same domain.

And then then something along the same lines for the second part.

Is this too cheap or incorrect somewhere? I'm kind of awful at rigor so any help would be greatly appreciated.
Yes, "And then then something along the same lines for the second part." does lack rigor.

What you have for the first part looks good. The non-rigorous second part may be OK.

... But, it's an 'if and only if ' theorem, so you're half done at best.

[Elwin.Martin]

Yes, "And then then something along the same lines for the second part." does lack rigor.

What you have for the first part looks good. The non-rigorous second part may be OK.

... But, it's an 'if and only if ' theorem, so you're half done at best.

^^; I wasn't proposing the second part as an actual part of a proof, haha. Thanks for reminding me about the need to proof the if and only if in both directions, though!


Since f is a set of ordered pairs and g is a set of ordered pairs, for f = g we must have that every ordered pair in f is an ordered pair in g and every ordered pair in g is an ordered pair in f (subsets of one another). For f and g to be equal then, they must contain the same first members and thus the same domain.
Since f is a set of ordered pairs and g is a set of ordered pairs, for f = g we must have that every ordered pair in f is an ordered pair in g and every ordered pair in g is an ordered pair in f (subsets of one another). For f and g to be equal then, they must contain the same pairing of y to each x in (x,y), i.e. they must have that for every x we have f(x)=g(x).

Suppose f and g are two sets of ordered pairs such that the first members of f are the same as the first members of g (same domain) and that for every x in the domain of f, f(x) = g(x). We would like to verify that f and g are equal, i.e. that they contain they are the same set of ordered pairs. Since f and g have the same first members, they share the same domain. To be functions then they must each have that (x,y) = (x,z) iff y=z; taking the fact that f(x) = g(x) for all x in the domain of f, and hence the domain of g, we have then that (x,f(x))=(x,g(x)) for all x in the domain of f and g. Thus, f(x)=g(x).

Is that complete? I feel like it's too wordy...the problem with these kind of proofs is that they feel self evident sometimes and it seems like so much effort explaining it, though I wouldn't doubt there's a small flaw in there.