Suppose for an example like S={-1,0,1}. I noticed that if you can only add two different elements it would remain closed however if you add the upper or lower bound to itself it would be out of the range of the set. Wouldn't closure under addition just fail hold for all finite sets of real numbers? (other than empty set or 0 set) I don't get what I'm trying to prove here.
2026-04-03 08:44:05.1775205845
Proving sets are closed under addition
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For any finite set of real numbers, we cannot be sure that always the closure under will not hold true. What you said holds true as long as the result(of the addition of two members of the set) is not part of the set.
However, if we take the case of just the upper, lower bounds with a finite set then we can be sure that the closure will not hold true.