Suppose 1 weren't an upper bound.

Then there exists some such that .

Thus, , particularly .

So 1 must be an upper bound of .

So is a bounded set. We still do not know if thereisa supremum in the rationals because , and does not carry the least upper bound property.

However, we know since the realsdohave the least upper bound property that at least lies in .

Suppose 1 were not the least upper bound of .

Let .

Then .

By density of the rationals in the reals, there exists such that .

Since , then is an upper bound of .

Since , write , for . Note that , for if not then which cannot be true.

Then .

However, if this were true, then would not be an upper bound of because for , then .

So therefore cannot be an upper bound of (and certainly not its least upper bound).

Therefore, the least upper bound of is 1 as desired. QED.