Consider $X = \{y \in \mathbb Q| y^2 <2\}$
Now I am a bit confused by the following:
It is true that $X$ is bounded by above as numbers such as $1.5$ are upper bounds. However we say it has no supremum.
Why is that the case? Had the set been defined as $X = \{y \in \mathbb Q| y^2 ≤ 2\}$ would then it have had a supremum? Any help would be appreciated.
The sets $\{y \in \Bbb Q \mid y^2 < 2\}$ and $\{y \in \Bbb Q \mid y^2 \leq 2\}$ are precisely the same set, since there is no $y \in \Bbb Q$ such that $y^2 = 2$.
The set $X$ does not have a rational least upper bound. Given any rational upper bound (such as 1.5), there exists a lower rational upper bound (such as 1.42).
The set $X$ does have a real supremum, however. In particular, $\sqrt{2}$ is a real number and is the least upper bound of $X$. That is, there is no $y \in \Bbb R$ such that $y < \sqrt{2}$ and $y$ is an upper bound for $X$.