# A simple question that I don't know what I should call it

1. Sep 6, 2011

Assume f:A->B is a function. Can we show that x$\in$A iff f(x)$\in$f(A) using only the axioms of Zermel-Fraenkel set theory and the definition of a function?

the statement is intuitive and obvious. but I want to know if it can be proved rigorously using the axioms and the definitions we have.

Last edited: Sep 6, 2011
2. Sep 6, 2011

### chiro

With regards to f(x) is an element of f(B), that doesn't make sense. The function is a mapping that is sending elements of A to B not A to f(B). I'm not trying to be a math nazi, just trying to help you.

If you are only given that all points (all evaluations of the function) are a subset of B, without the function definition itself, then you can't really say what space the domain came from. For example if B was the real line, you could have a simple one variable function like y = x^2, or you could have the dot product in 3D space that sends two vectors to a real number. Given B without a definition does not allow you to deduce the domain.

If you are given the function definition however, then it seems to be a moot point. You could use just standard set theoretic tools to show that the range of the function is actually B and that the codomain is some subset.

I'm not quite sure though what you're trying to do, so I could be off topic.

3. Sep 6, 2011

### disregardthat

Well, f(x) doesn't make sense unless x is in A, so yes, its true, since "f(x) is in f(A)" is a tautology (true for all functions) and assumes x is in A.

4. Sep 7, 2011

f(x) is an element of f(A).

You're missing the main point of the question I guess. we know that if a=b then f(a)=f(b). the converse is not necessarily true unless f is one-to-one. now suppose x$\in$A. can we do the same thing here and say f(x)$\in$f(A)? intuitively, the answer is yes. What about the converse? if f(x)$\in$f(A) can we say that x$\in$A? I suppose the answer is yes when f-1 is defined.

How do you know that?
Again, How do you know that?
your argument is a type of petitio principii.

I don't want to know if the statement is true or not, the statement itself is too obvious that I want it to be clarified for me. I want to know if we can prove it using a set of axioms and the definition of function. one may think that my question is stupid, it really is! but we can't assume something is true unless it is derived from the definition of another thing related to it or we prove it using tautologies and axioms.

5. Sep 7, 2011

### disregardthat

No, f is a function from A to B, so f(x) does not make sense unless x is in A. Compare: Define f: N --> N by f(n) = 2n. What is f(0.4) ? You can't talk about f(0.4), since 0.4 is not in N.

6. Sep 7, 2011

That doesn't prove anything.

Any ideas how to prove it axiomatically?

7. Sep 7, 2011

### pwsnafu

You certainly don't need to specify a formula. The domain is simply "dom f".

Of course, its not that useful, but we are not discussing that.
And it is irrelevant to OP's question anyway.

OP,
the set f(A) is by definition the image of A under f, that is in symbols
$f(A) := \{ b \in B \, | \, \exists x \in A, \, f(x) = b \}$
This is why disregardthat is calling it a tautology. There is nothing to prove.

8. Sep 7, 2011

Thanks. That sounds convincing. so if $x \in A$ & $f(x)=b$ then $b=f(x) \in f(A)$. conversely, if $b=f(x) \in f(A)$ then $\exists x: x \in A$.
Is that right?

9. Sep 7, 2011

### chiro

You guys are right, I didn't really get what the OP was asking for.

It sounded like a circular kind of question, and by the sounds of the responses it kind of was.

10. Sep 7, 2011

### pwsnafu

I think both disregardthat and I were going a bit too fast. Let's do this ridiculously slowly.

First, left to right.
Suppose f is a function with domain A, codomain B and graph Γ.
Then suppose that $$x \in A$$
By definition of the symbol "f(x)" we know there exists a pair $$(a,b) \in Γ$$ such that $$a = x$$ and $$b = f(x)$$
Now $$f(A) = \{ b \in B | (a,b) \in Γ \}$$
Then so $$b = f(x) \implies f(x) \in f(A)$$
QED.

Second, right to left.
We do a contrapositive.
Assume $$x \notin A$$
Then because $$Γ \subset A \times B$$, there does not exist $$(a,b) \in Γ$$ such that $$a = x$$
But b is f(x). So any pair of form $$(c, f(x))$$ cannot be an element of Γ
And $$Γ \subset A \times B \implies f(x) \notin B$$
Lastly $$f(A) \subset B \implies f(x) \notin f(A)$$
QED.

Last edited: Sep 7, 2011