Proving that $A+A \not\subset 2A $ using a particular definition for sum of sets

Question

Let $X$ de a vector space and $A\subset X$. If we define $A+B :=\{a+b|a\in A, b\in B\}$ I am trying to show that

$$ \begin{equation} 2A \subsetneq A+A, \end{equation} $$

i.e., it might happen that $2A \neq A+A$. I was able to show that $2A \subset A+A$, but I am not able to come up with a counterexample to show that the converse is not necessarily true. If I assume $A+A\subset 2A$ and let $x\in(A+A)$, then by its definition we have $x=a_1 + a_2$ for some $a_1,a_2\in A$ I guess I could naively say that $a_1+a_2 = 2(a'_1+a'_2)$ where $a_i=a_i/2$ for $i=1,2$ and I guess $(a'_1+a'_2)$ might not be an element of $A$ in some concrete case but I am not able to come up with anything and I've spent more time than I would like to admit. Any suggestions?

Thank you.

Answer 1

When you are looking for counter-examples, it is often a good idea to start with small sets. In the context of vector spaces, you might start with sets that contain only one of the identity elements, or an identity element and something else. For example,

• First try: $A = \{ 0 \}$ (i.e. $A$ contains only the additive identity). Then $A+A = \{ 0 \}$ and $2A = \{ 0\}$. Nope. That doesn't work.
• Second try: $A = \{ 1 \}$ (i.e. $A$ contains only the multiplicative identity). Then $A+A = \{ 2 \}$ and $2A= \{ 2 \}$. Nope again.
• Third try: $A = \{ 0, 1\}$. Then $A + A = \{ 0, 1, 2\}$ and $2A = \{ 0, 2\}$. Oh! This gets the job done! Yay!

Again, when you are looking for counter-examples, it is good to start simple. That won't always work, but it is a good way to get started.