Proving if $|A|\ge 4 \vee |A|\le 2$ then $|A+A|\neq 4$ with direct, contradiction and contraposition

Question

Prove if $|A|\ge 4 \vee |A|\le 2$ then $|A+A|\neq 4$. $A$ is some set and we define $A+B=\{a+b|a\in A, b\in B\}$, $A$ is some subset of the reals.

In a direct proof and proof by contradiction I'd have to check every case for $|A|=0,1,2,4$ right?

In a proof by contra position I'd have to show that if $|A+A|=4$ then $|A|=3$ but I'm getting a contradiction, suppose $A+A=\{a,b,c,d\}$ then

$a=m+m\\b=n+n\\c=l+l\\d=m+n$

But there are still two elements left: $n+l,m+l$. How could it be that a direct proof works but a contra position results in a contradiction?

Answer 1

I think it's easier to prove this directly. If |A| = 1, it's clear that |A + A| is not 4. If |A| = 2, then A + A can have at most 3 elements. If |A| = n > 4, then let the elements of A be:

a1, a2, ...., an

We know that a1, ..., a5 are distinct. So 2a1, 2a2, ..., 2a5 are distinct, but these are in the form ai + ai with ai in A; thus, A + A has at least 5 elements.

It remains to handle the case |A| = 4. For this, let A = {a1, a2, a3, a4}. It's easy to see |A + A| >= 4 because 2ai are distinct, for all i. Now if |A + A| = 4, then it must be that:

A + A = {2a1, 2a2, 2a3, 2a4}.

Now, work out a contradiction by forcing the existence of an additional element ak + at for some k and t between 1 and 4.

Answer 2

The rules of the game are unclear.

You have to tell us the underlying (abelian) group of which $A$ is a subset. E.g., if $A={\mathbb Z}_2\oplus{\mathbb Z}_2$ is the four group then $|A|=4$ and $|A+A|=|A|=4$.