Can anybody help me with this? The question is:
Let $S$ be a subset of $\mathbb{R}$ such that $S \not = \emptyset$ and $S \not = \mathbb{R}$. Define $S^c =\{x\in \mathbb{R}:x \not \in S\}$. Suppose that $S$ is closed under addition AND closed under the additive inverse. Prove $S^c$ is not closed under addition. There is an example where $S$ and $S^c$ are both closed under multiplication.
My attempt at the proof is:
I will prove this directly. That is, I will assume choose two elements $x,y \in S^c$ and show that $x+y \not \in S^c$. This means, I will have to show $x+y \in S$. We know $S$ is a subset of the real numbers that is not the empty set nor the set of real numbers. I will choose elements $x$ and $y$ in $S^c$ and add them to each other to get $x+y$. We will let $y$ be $-x$ so we get $x+(-x)$ which we know is 0 by the definition of additive identity. Since 0 is not in the empty set and it is in a subset of $\mathbb{R}$, we know 0 is in $S$ therefore proving $S^c$ is not closed under addition.
We will now provide an example where $S$ and $S^c$ are both closed under multiplication. $S$ and $S^c$ are both closed under multiplication when there is an element $y \in S$ such that $y=0$ and there is an element $x \in S^c$ such that $x=0$. This is because anything multiplied by zero is zero by a proof proven previously in class. Thus we know $S$ and $S^c$ are both closed under multiplication in this case which concludes the proof.
I am pretty sure this is incorrect in many ways, so I was wondering if anyone can help me?