Mentor

## (x,x] = {x}

 Quote by SW VandeCarr The interval (b,a] is open to b (the interval doesn't include b) but closed in 'a' (the interval includes a.)
If you define (b,a] as "the interval from b to a, with b not included and a included", then (x,x] is neither empty nor non-empty, it's just nonsense. (A set can't both contain x and not contain x). If you define (b,a] as "the set of all t such that b<t≤a", then (x,x] is empty. (I prefer the latter definition because it makes sense for all a and b, and doesn't require us to have defined "interval" in advance).

 Quote by SW VandeCarr I also don't understand why you were talking about countable sets. It seems to me the notation (x,x] implies an interval on the real number line unless otherwise specified.
I don't understand your concern at all. n is an index that labels the sets in the family of sets that we're taking the intersection of. If the definitions of the sets in the family hadn't involved some algebraic operations, we could have used any set as the index set. In this particular case, where the sets are (x-1/n,x], the index set can be any subset of the real numbers that doesn't include 0.

 Quote by tiny-tim why should it? the interval (3,2] doesn't contain 2, so why should (2,2] ? yes, elements "strictly greater than x and less than or equal to x" don't include x
I'm sorry. Why do you say that (3,2] doesn't include 2? As a half closed interval doesn't this notation correspond to $$3> x \geq 2$$? Please explain why you are saying 2 is not included in the interval when every reference I've checked says it is. I'll just post one.

http://mathworld.wolfram.com/Half-ClosedInterval.html
 Mentor No, (3,2] is the set of all real numbers x such that 3

 Quote by Fredrik No, (3,2] is the set of all real numbers x such that 3
OK. But that's not the inequality I wrote in my previous posts. However for (a,b] I do see your point. If I write $$a < x \leq b$$ then the corresponding set has no least element and therefore cannot be well ordered. Thanks.