# Find the supremum $(0,1)\setminus \mathbb{Q}$.

My intuition tells me that $A = (0,1)\setminus\mathbb{Q}$ has no supremum. But I'm not sure if the following arguments are sufficient or if my intuition is misguided...

We have that 1 is an upper bound of A, since:$$1 > a \quad \forall a \in A$$

then because $\mathbb{Q}$ is dense in $\mathbb{R}$, there exists a $q$ in $\mathbb{Q}$ such that: $$a < q < 1 \quad \forall a \in A$$

hence we can find an $r$ in $\mathbb{Q}$ such that $$a < r < q \quad \forall a \in A$$

and the process repeats indefinitely, hence we will never find a least upper bound for the set. I assume I can use something similar to prove that $A$ has no infimum...

Any help is appreciated!

• What do you mean by $(0,1)/\mathbb{Q}$? – carmichael561 Mar 14 '17 at 1:00
• That is the way I have it in my homework assignment. I guess it means all the irrational numbers in the interval $(0,1)$ – migueldva Mar 14 '17 at 1:02
• @MathematicsStudent1122: Why the hostility? The OP posted a question and his attempt to solve. – Matthew Leingang Mar 14 '17 at 1:09
• @MathematicsStudent1122 I guess you can then show us your brilliance and prove it. – migueldva Mar 14 '17 at 1:11
• Note that the supremum is the least upper bound of a set.This does not say that supremum is a member of said set,it can be outside of the set. If you like you say that the process of trying to find an upper bound repeats endlessly then what does that say about what an upper bound may be? – Mathew Duxbury Mar 14 '17 at 1:21

We have that 1 is an upper bound of A, since:$$1 > a \quad \forall a \in A$$

Yes, this is right.

then because $\mathbb{Q}$ is dense in $\mathbb{R}$, there exists a $q$ in $\mathbb{Q}$ such that: $$a < q < 1 \quad \forall a \in A$$

Not so fast. Yes, $\mathbb{Q}$ is dense in $\mathbb{R}$, but you have your quantifiers mixed up. It is true that for all $a \in A$ there exists $q \in \mathbb{Q}$ such that $a < q < 1$. What you wrote is that there exists $q \in \mathbb{Q}$ such that for all $a \in A$, $a < q < 1$. Do you see the difference? The second (false) statement asserts that there is an upper bound for $A$ less than $1$. The first (true) statement asserts that no $a\in A$ is maximal.

I think what you mean to say is that for all $x\in\mathbb{R}$ with $x<1$, there exists $a\in A$ such that $x < a < 1$.

• Got the difference. Thanks! – migueldva Mar 14 '17 at 1:20

The set $A$ is non-empty and bounded above by $1$. It therefore has a supremum $\sup A \le 1$ in $\mathbb{R}$. You can also verify that $$1-\varepsilon < \sup A,$$ for all $\varepsilon > 0$. Therefore $$\sup A = 1.$$

• Your "you can also verify that..." neatly obscures the only part of the proof that requires any work. – carmichael561 Mar 14 '17 at 1:13
• Not a lot of work, but thank you. :-) – Olivier Mar 14 '17 at 1:15

I suspect your problem is this: you think that the supremum of a set $S$ must belong to $S$. This is not true! If you keep that in mind, I think you will easily see what the supremum must be.

By the way, the supremum is different from the maximum: the maximum of a set $S$ does have to belong to $S$. So not every bounded set has a maximum, although every bounded non-empty set has a supremum.

• Yes. Thank you! – migueldva Mar 14 '17 at 1:20