# When defining closure property of the set $\{0,1,-1\}$ under addition? Can we say that this set is closed under addition?

When defining closure property of the set $$\{0,1,-1\}$$ under addition? Can we say that this set is closed under addition?

My confusion is that if we take pairs like $$(0,1),(0,-1),(1,-1)$$ then the seems to be closed under addition. Can we take $$1$$ two times and say $$1+1=2$$ and as $$2$$ doesn't exist in the set, hence the given set is not closed under addition. Please help me in this.

• Yes, that set is not closed under addition because $1+1=2$ is not in it – J. W. Tanner Apr 30 at 18:58
• but a set contains $1$ element only once, how we can take same element twice from it – prat Apr 30 at 19:01
• The definition of closure of a set $S$ under an operation $\oplus$ says that for any elements $s, t\in S$, $s\oplus t\in S$. It doesn't say that $s$ and $t$ have to be distinct elements. – Steven Stadnicki Apr 30 at 19:04
• But the set is closed under usual multiplication. – Pritam Apr 30 at 21:23