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I'm having a bit of trouble understanding this exercise:

Find proper subsets $A$, $B$, $C$ of a vector space $V$ over the field $\mathbb{R}^2$, so that $A + A \subset A$, $B \subset B + B$ and $C + C = C$.

Questions: Can I define the operation $+$ myself (f.e difference, or union etc) as long as $V$ still remains a vector space?

If yes: Since $A$, $B$, $C$ are subsets of the same vector space $V$, I can not define 3 different operations, right? One operation must be able to handle all those cases?

If yes: How is this possible?

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What do you mean by the field $\Bbb R^2$? I expect that for $A,B\subseteq V$, what is meant by $A+B$ is $\{a+b:a\in A\text{ and }b\in B\}$, so no, you probably don’t have any choice about what $+$ is. – Brian M. Scott Mar 29 '12 at 23:38
Does $A=0$, $B=0$, $C=0$ work? – Peter Grill Mar 29 '12 at 23:39
@Dylan, perhaps one is intended to interpret the inclusions as strict. – Gerry Myerson Mar 29 '12 at 23:41
I bet you can even find proper, non-trivial subsets $A,B,C$ of the integers such that $A+A$ is a proper subset of $A$, $B$ is a proper subset of $B+B$, and $C+C=C$. – Gerry Myerson Mar 29 '12 at 23:43
Once again, hooray for $\subseteq,\subsetneq$ and for the extra joyous: $\subsetneqq$. – Asaf Karagila Mar 29 '12 at 23:44
up vote 1 down vote accepted

You're not defining a new operation, you are using the usual $+$ operation of $\mathbf{V}$.

If $\mathbf{V}$ is a vector space (or more generally, an abelian group), and $S$ and $T$ are subsets (not necessarily subspaces) of $\mathbf{V}$, then by definition $$S+T = \{s+t\mid s\in S,\ t\in T\}.$$

For $A$, you are looking for a subset with $A+A\subset A$, $A+A\neq A$.

  • Hint. If $\mathbf{v}\neq\mathbf{0}$, and $\mathbf{v}\in A$, then $2\mathbf{v}=\mathbf{v}+\mathbf{v}\in A+A\subset A$, so $3\mathbf{v}=2\mathbf{v}+\mathbf{v}\in A+A\subset A$, etc. So you will at least have $\{ n\mathbf{v}\mid n\text{ a positive integer}\}$ in $A$. Can that be all you have in $A$?

For $B$, you want a subset with $B\subset B+B$, $B\neq B+B$.

  • Hint. Well, if you have $\mathbf{v}\in B$ as above, but $2\mathbf{v}\notin B$, then that will guarantee they are not equal. But $B=\{\mathbf{v}\}$ will not work, because then $\mathbf{v}\notin B+B = \{\mathbf{v}+\mathbf{v}\} = \{2\mathbf{v}\}$. Maybe you can add stuff to $B$ to guarantee $\mathbf{v}\in B+B$? Careful, though, because now you'll need the thing you added to also be in $B+B$...

For $C$ you want something with $C+C=C$.

  • Hint. What happens if you have all of $\mathbf{V}$?
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Thanks, now that I know how + is defined, I can solve the rest. – gq3 Mar 30 '12 at 11:11

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