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.

  • $\begingroup$ Yes, that set is not closed under addition because $1+1=2$ is not in it $\endgroup$ – J. W. Tanner Apr 30 at 18:58
  • $\begingroup$ but a set contains $1$ element only once, how we can take same element twice from it $\endgroup$ – prat Apr 30 at 19:01
  • 3
    $\begingroup$ 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. $\endgroup$ – Steven Stadnicki Apr 30 at 19:04
  • $\begingroup$ But the set is closed under usual multiplication. $\endgroup$ – Pritam Apr 30 at 21:23

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