Question regarding partially ordered sets I have encountered few questions while reading the book 'Modern Algebra'.
Let $\mathbb Q$ be the set of rational numbers.
Let

$B = \{ x : x\in\mathbb Q,\sqrt2 < x < \sqrt3 \}$.

How it can be shown that -

*

*$B$ has infinite number of upper and lower bounds.


*$\inf B$ and $\sup B$ do not exist .
Why $\sqrt2$ and $\sqrt3$ cannot be taken as lower aqnd upper bounds?
I am not able to understand that how can infimum and supremum can exist at the first place as the relation defined is not a binary relation.
A detailed proof would be helpful.


Why $\sqrt2$ and $\sqrt 3$ can not be taken as lower and upper bounds?

Because the exercise is implicitly asking about an inf and sup that are in $\mathbb Q$. The exercise is perhaps not stating this very clearly, but remember that infimum and supremum is always about how a set fits into a certain ordered superset.

I am not able to understand that how can infimum and supremum can exist at the first place as the relation defined is not a binary relation.

The order relation that is being considered is the ordinary comparison of rational numbers. $x\le y$ is certainly a binary relation. (This is the default assumption when you speak about order properties of sets of numbers).
