Thread: Supremum not an element of set, prove infinite

Let A be a nonempty subset of the Reals bounded above. Suppose sup A exists and that sup A is not an element of A. Show that A contains a countably infinite subset. In particular, A is infinite.

I'm reasonably sure that the easiest way to prove this is using an epsilon argument, but I don't really know where to start on here.

Here's my scrambled thoughts:

Let a = sup A
x be an element of A
x <= a

For any epsilon > 0 there is a-epsilon < x < a
how do a I show that the interval from a-epsilon to x is countably infinite?

Thanks

Originally Posted by jsmith90210

Let A be a nonempty subset of the Reals bounded above. Suppose sup A exists and that sup A is not an element of A. Show that A contains a countably infinite subset. In particular, A is infinite.

I'm reasonably sure that the easiest way to prove this is using an epsilon argument, but I don't really know where to start on here.

Here's my scrambled thoughts:

Let a = sup A
x be an element of A
x <= a

For any epsilon > 0 there is a-epsilon < x < a
how do a I show that the interval from a-epsilon to x is countably infinite?

Thanks

I think the best way to do this is to do the easy proof that if with then and so . So, for one MUST have that is not finite. Sound feasible? Or do you want to do the more direct proof?

I see the logic in your proof, but if possible, I'd like a more direct proof

Originally Posted by jsmith90210

Let A be a nonempty subset of the Reals bounded above. Suppose sup A exists and that sup A is not an element of A. Show that A contains a countably infinite subset. In particular, A is infinite.

Suppose .
WHY?

If then . WHY?

How does that prove it?

Should that be xn-1 < xn? And I don't know how that proves it, that's the part I'm struggling with

Originally Posted by jsmith90210

Should that be xn-1 < xn? And I don't know how that proves it, that's the part I'm struggling with

1) if then it is not an upper bound. So?

2) we have
That is an infinite collection.