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I'm trying to find an example of a nonempty subset of $\mathbb{R}^2$ that is closed under addition and additive inverses but is not a subspace of $\mathbb{R}^2$.

I know the $0$ vector has to be in the subset since its closed under addition and additive inverses, but I can't think of anything that fails to hold under scalar multiplication. Any ideas?

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3 Answers

Consider the set $\Bbb Z\times \{0\}=\bigl\{\langle x,0\rangle\in\Bbb R^2:x\in\Bbb Z\bigr\}.$

More generally, take any $x_1,x_2,y_1,y_2\in\Bbb R$ with $x_1,y_1$ not both zero, $x_2,y_2$ not both zero, and such that the line in $\Bbb R^2$ through $\langle x_1,y_1\rangle$ and $\langle x_2,y_2\rangle$ doesn't pass through the origin. Then $$\bigl\{\langle ax_1,ay_1\rangle\in\Bbb R^2:a\in\Bbb Z\bigr\}\tag{1}$$ and $$\bigl\{\langle ax_1+bx_2,ay_1+by_2\rangle\in\Bbb R^2:a,b\in\Bbb Z\bigr\}\tag{2}$$ are additively closed sets with additive inverses.

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Hint: What are the analogues of $\mathbb{Q}, \mathbb{Z}\subseteq\mathbb{R}$ as subsets of $\mathbb{R}^2$?

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There are many examples, but here is a more general method:

  1. Take any countable subset $A_0 \ni 0$ of $\mathbb{R}^2$.
  2. Set $f(X) = X \cup \{p-q \mid p,q \in X\}$.
  3. Construct $A_{n+1} = f(A_n)$.
  4. $A = \lim_{n\to\infty}A_n$ will be your solution.

The limit exists by Knaster-Tarski fixed point theorem, and is a fixed point of $f$, that is

$$f(A) = A$$

and thus is closed under addition and additive inverses. However, $|A| \leq \aleph_0 < 2^{\aleph_0}$ and thus cannot be a subspace of $\mathbb{R}^2$ (countable union of countable sets is still countable).

Similar approach would also work for some uncountable seeds, but then you have to prove, that the result is not a subspace of $\mathbb{R}^2$.

I hope this helps ;-)

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