# How can you prove this set theory statement by contradiction [duplicate]

$$f(A\cup B)=f(A)\cup f(B)$$

My attempt:

Suppose that $f(A\cup B) \neq f(A) \cup f(B)$, then $f(A\cup B)$ has a variable $x$ that is neither in $A$ nor $B$; however, $A\cup B$ implies that $x$ must me in either $A$ or $B$, thus we have a contradiction.

Yeah... it's terrible, but I have no intuition as to how to prove this in any other way. Also, the proofs I saw in the lectures were sometimes as unconvincing as this one I just came up with.

It's like they're saying this is this because it is so. How do you develop the skills and intuition to solve these kind of problems?

Also, you have to know it is true before proving it. How do you even know it is true just by looking at it?

-

## marked as duplicate by Claude Leibovici, Daniel Robert-Nicoud, AlexR, Lost1, TMMFeb 1 '14 at 14:00

Are you implying that it's not possible? – Gladstone Asder Oct 26 '12 at 4:42
@GladstoneAsder Oftentimes to prove two sets are equal, mutual set inclusion is the easiest way. Contradiction is sometimes unnatural like in this case (though certainly possible). – EuYu Oct 26 '12 at 4:43
If the proofs you have seen in lectures are as unconvincing as the "proof" you give, then either you have the world's most incompetent lecturer or, more likely, your evaluation of the convincingness of an argument requires major recalibration. You write "variable" when there are no variables; you write of a set implying something, which is not something sets do; and so on. You develop the skills by analyzing correct arguments until you understand why they are correct, and by paying close attention to definitions and correct usage of vocabulary. – Gerry Myerson Oct 26 '12 at 4:56

I think you've conveyed the general idea of the proof well enough, though the delivery could use some polishing and you are missing quite a bit of detail. I don't really blame you though since a proof by contradiction in this scenario is a bit unnatural.

Suppose for the sake of contradiction that $f(A) \cup f(B) \neq f(A\cup B)$. Then there either exists $x\in f(A) \cup f(B)$ such that $x\notin f(A\cup B)$ or $x\in f(A\cup B)$ such that $x\notin f(A) \cup f(B)$. We will handle these separately.

First suppose that $x\in f(A) \cup f(B)$ and $x\notin f(A\cup B)$. Then $x\in f(A)$ or $x\in f(B)$. We must then have $x\in f(A) \subseteq f(A\cup B)$ or $x\in f(B) \subseteq f(A\cup B)$. A contradiction either way.

Suppose then that $x\in f(A\cup B)$ and $x\notin f(A) \cup f(B)$. Then there exists $y \in A\cup B$ such that $f(y) = x$. We have $y\in A$ or $y\in B$. In either case, $f(y) \in f(A)\cup f(B)$ and we obtain a contradiction.

If you have any experience with mutual set inclusion proofs then you'll see that the above is essentially a convoluted way of writing mutual set inclusion. The contradiction part is completely unnecessary and adds nothing to the proof.

A bit of a remark on your last question. Certainly the statement has to be true for you to prove it. A proof is a certificate of correctness. But oftentimes you have no idea as to whether a given statement is true or false.

In textbook scenarios, you are generally given a problem - guaranteed to be true - for you to prove. More often than not, the statement is not only true but has an elementary and elegant solution (i.e. olympiads). This is in direct contrast to say, mathematical research, where not only are the statement given not guaranteed to be true, but you often struggle to find even an underlying framework to present your problem. Research is done often with simultaneous approaches in proving and finding counter-examples to any given problem.

The task of deciding the validity of a statement ultimately falls upon your intuition, and human intuition is developed more for hunting buffalo than for judging mathematical validity. So as a rule of thumb, never assume anything to be true unless you have a valid proof of the statement, there have been cases in history in which we have been burned quite badly by our intuition.

-

I won't do a proof by contradiction, but I'll show you how a direct proof might go. If you study it closely, you'll see that all I'm doing is unraveling the definitions and following my nose.

Suppose $x$ is in $f(A\cup B)$. That means there exists $c$ in $A\cup B$ such that $f(c)=x$. Now $c$ in $A\cup B$ means $c$ is in $A$ or $c$ is in $B$. If $c$ is in $A$ then $x$ is in $f(A)$. If $c$ is in $B$ then $x$ is in $f(B)$. Since $c$ is in $A$ or $c$ is in $B$, $x$ is in $f(A)$ or $x$ is in $f(B)$. So, $x$ is in $f(A)\cup f(B)$. We have proved $$f(A\cup B)\subseteq f(A)\cup f(B)$$

Can you do the inclusion in the other direction, thus completing the proof?

-