Least Upper Bound Property. I had a question about the Least Upper Bound Property. So it states that every non-empty subset of $\mathbb{R}$ that has an upper bound must have a least upper bound in the reals.
My question is: Does the set of all real irrational numbers with simple order have the least upper bound property?
I was thinking about this and said no. This set isn't necessarily bounded (at least I don't see how it's bounded) which is why I don't think it has the least upper bound property. But someone was telling me this set has the least upper bound property and their reason was it's a subset of the reals and every subset of the reals has a least upper bound property.
So I'm not sure if I'm right or my friend is right, but I feel like they're wrong which is why I was wondering.
 A: This is kind of ambiguous depending on what you mean by "$X$ has the least upper bound property". The standard interpretation is: $X$ has the least upper bound property if every nonempty subset of X with an upper bound has a least upper bound in $X$. In this case, your friend is clearly wrong. Take $\mathbb{Q}$ as a counterexample: the set $\{x\in\mathbb{Q}: x^2\leq 2\}$ is bounded above, but it has no least upper bound in $\mathbb{Q}$. When this set is viewed as a subset of $\mathbb{R}$ though, it does have a supremum--namely, $\sqrt{2}$.
The set of all irrational numbers does not have the least upper bound property then either, since I can just define a set $\{x\in\mathbb{I}: x< 2\}$. But again, nonempty bounded subsets of $\mathbb{I}$ do have supremums in $\mathbb{R}$. For this set, the supremum is $2$.
A: The prerequisite is the set must has at least one upper bound. Technically the property should be stated like this: 
If there is a set $V$ such that:
1).$V\ne\varnothing$
2).$\exists a\in R$ such that $\forall x\in V$, $x\le a$
Then $\exists sup(V)\in R$ which is the least upper bound of $V$.
