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My linear algebra book (Linear Algebra Done Right by Sheldon Axler) has the following problem as exercise 1.6:

Give an example of a nonempty subset $U$ of $\mathbb{R}^2$ such that $U$ is closed under addition and under taking additive inverses (meaning $-u \in U$ whenever $u \in U$), but $U$ is not a subspace of $\mathbb{R}^2$.

It seems to me that such a set cannot exist, since the only subspace condition it's not mandated to fulfill is containing $0$, and for any $u \in U$, I can negate it to get $-u$ and then get $u + (-u) = 0$.

What is going on?

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  • $\begingroup$ $0$ is not the issue here. Try $\mathbb{Q}^2$. $\endgroup$ May 3, 2015 at 16:14

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Hint: Consider all ordered pairs $(a,b)$ where $a$ and $b$ are integers. Or, a little simpler, all ordered pairs $(a,0)$ where $a$ ranges over the integers.

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  • $\begingroup$ Ah, I forgot that $U$ needs to be a vector space in its own right. Thanks. $\endgroup$
    – Eli Rose
    May 3, 2015 at 16:23
  • $\begingroup$ You are welcome. Closure under scalar multiplication fails. $\endgroup$ May 3, 2015 at 16:35

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