I need to prove that the set of rational numbers in the closed interval 0,1 has a supremum and infimum. I know that they exist and I also know that I need to use proof by contradiction but I don't know how..
I'm really struggling with this so any help would be hugely appreciated!
thank you
On
Write down the definition of supremum and infimum. Not sure why you need to use a proof by contradiction, but it is relatively straightforward to show that $1$ is the supremum and $0$ is the infimum. This will hold even if you change your closed interval to an open interval.
Use the definitions.
Let $A = [0,1] \cap \mathbb{Q}$. If $a \in A$, then $a \in [0,1]$, so $0 \le a \le 1$. So $A$ is bounded above and below by $0$ and $1$ respectively. So we have that $0 \le \inf(A) $ and $ \sup(A) \le 1$.
Now if $a \in A$ and $a < 1$, then $a < \frac{1 + a}{2} < 1$, and it's easy to see that $\frac{1+a}{2} \in A$. This shows that $\sup(A) = 1$. Argue similarly on the infimum.