Let $A = ${$r \in \mathbb Q :\, \sqrt3 < r \le 5 \}$. What is the $\inf A$ and $\sup A$?

Question

Let $A = ${$r \in \mathbb Q :\, \sqrt3 < r \le 5 \}$. What is the $\inf A$ and $\sup A$?

So off the top of my head without writing a proof, I would assume that the $\inf A$ wouldn't be $\sqrt3$ because it is not a rational number. So I am not sure what the inf A should be. And then for the $\sup A$, I would assume $5$ is the $\sup A$ and that should be alright since it is a rational number in the set $A$.

For the proof though, I am unsure how to prove it. I am guessing I would have to show that whatever the $\inf A$, that it is the greatest lower bound and then for the $\sup A$ which I believe is $5$, that it is the least upper bound.

Answer 1

You have a fundamental misunderstanding of what $\inf$ means.

If $A$ is a set that is bounded below, there is nothing in the definition that says $\inf A \in A$. In fact, so often do we have $\inf A \not \in A$ that one might say that is the entire point of $\inf$.

The definition of $\inf A$ is that i) for every $a\in A$ we have that $\inf A \le a$. (In other words, $\inf A$ is a lower bound.) In this case $\sqrt 3 < a$ for all $a \in A$.

and ii) that $\inf A$ is the largest such number that is. That is if $y > \inf A$ then $y$ is not a lower bound. Or in other words if $y > \inf A$ there will be a $b\in A$ so that $\inf A \le b < y$.

For example $1$ is a lower bound of $A$ because if $\sqrt 3 < a \le 5$ then $1 < a$. But $1 \ne \inf A$ because $1.2 > 1$ and $1.2$ is also a lower bound so $1$ is not the greatest lower bound.

Now what about $\sqrt 3$? If $y > \sqrt 3,$ then can we find a rational number $b\in A$ so that $\sqrt 3 < b < y$? I mean if $y= 1.8$ then we can have $b = 1.74$ and if $y = 1.733$ we can have $b=1.7321$ and so on.

Is there any $y > \sqrt 3$ where $y$ is a lower bound?

If not, then $\sqrt 3$ is the greatest lower bound.

That $\sqrt 3$ is not rational and $\sqrt 3 \not \in A$ is completely irrelevant.

Answer 2

Although it is not stated explicitly in the question, the usage of symbol $\sqrt{3}$ suggests that you are working in the system of real numbers. In that case as you have rightly guessed $\inf A=\sqrt{3},\sup A=5$.

In case you wish to deal with this problem within $\mathbb{Q} $ then you should state this explicitly and avoid usage of symbol $\sqrt{3}$. The set $A$ can then be defined as $$A=\{x\mid x\in\mathbb {Q}, 0<x\leq 5,x^2>3\}$$ And then it is trivial that $\sup A=5$ but not so easy to prove that set $A$ does not possess an infimum in $\mathbb {Q} $.