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Proof-trying. Note that for all $x,y$, $x-y=x+(-y)$. Assume $X\subseteq\mathbb{R}$ is closed under subtraction, i.e., $X-X\subseteq X$. Thus, since $X-X=X+(-X)\subseteq X$, $X$ is closed under addition.

Can you give a hint for proof?

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    Where is the proof that you are asking us to check? – 2017-02-06
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    **Hint.** $x+y = x-((x-x)-y)$. – 2017-02-06

4 Answers 4

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If you know that $X=-X$, and $X+(-X)=X-X$, then $$X+X=X+(-X)=X-X \subset X$$

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    But $X=-X$ is not a hypothesis. – 2017-02-06
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    But it can be proven that if X is closed under subraction then X = -X. i.e. if $y \in X$ then you can prove $-y \in X$ (but you can't assume it). – 2017-02-07
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Your proof is not correct. To prove that $X$ is closed under addition, you need to prove that $X+X\subseteq X$. Instead you have proven that $X+(-X)\subseteq X$. To fix your proof, you might try proving that $X=-X$.

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No, your proof is incorrect.

A better proof is as follows:

If $X$ is empty, then it is closed under addition vacuously.

Since $X$ is closed under subtraction, if $x \in X$ (so $X$ is not empty), then $x - x = 0 \in X$ is also true.

Furthermore, for all $x \in X$, since $0 \in X$, it follows that $0 - x = -x \in X$.

Now, we can safely say: let $x, y \in X$, then $x + y = x- (-y) \in X$ since $-y \in X$ also, and because $X$ is closed under subtraction.Hence, we are done.

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Let $x, y \in X$ then $y-y = 0 \in X$. Then $0 - y = -y \in X$ then $x -(-y) = x+y \in X$.