Must there be a supremum in a bounded infinite set?

[The Student]

Must there be a supremum in a bounded infinite set? Why or why not?

*I am sorry, but I meant to put "Must a bounded infinite set have a supremum?" and not necessarily a member of the set.

 

[daon2]

If it is a bounded subset of real numbers, then yes. The set may or may not contain the supremum, of course.

And why? Well it is a result of the completeness property of the real numbers. There is a proof using Cauchy Sequences on wikipedia: http://en.wikipedia.org/wiki/Least-upper-bound_property

 

[The Student]

If it is a bounded subset of real numbers, then yes. The set may or may not contain the supremum, of course.

And why? Well it is a result of the completeness property of the real numbers. There is a proof using Cauchy Sequences on wikipedia: http://en.wikipedia.org/wiki/Least-upper-bound_property

If it is a bounded set of real numbers, does that mean that it has to be a subset?

 

[daon2]

If it is a bounded set of real numbers, does that mean it is a subset?

Of course

 

[The Student]

Ok, thanks!

 

[HallsofIvy]

Note that "every bounded set of real numbers has a supremum" (and an infimum- technically, "bounded" means bounded both above and below) is a defining property of the real numbers (the "completeness property"). It is NOT true of the set of rational numbers. The set of "all rational numbers whose square is less than 2" has, say, 1.5 as upper bound but does not have a supremum in the rational numbers. As a set of real numbers its supremum is \(\displaystyle \sqrt{2}\) but that is not a rational number.