# supremum ,infinum ,maximum,minimum

• Aug 2nd 2011, 02:39 AM
lukeheselden
supremum ,infinum ,maximum,minimum
i was doing a question and i got stuck again,
{x is an element of rational numbers: x^3 ≤ pi}
i think the answers are sup:cube root of pi,inf: no infimum ,max: cube root of pi, min:no minimum

am i right?
• Aug 2nd 2011, 03:04 AM
pece
Re: supremum ,infinum ,maximum,minimum
Remember, a maximum of a set $\displaystyle A$ is a element of $\displaystyle A$ such that every element of $\displaystyle A$ are smaller or equal to it.

So, is $\displaystyle \sqrt[3]{\pi} \in \{q \in \mathbb Q / q^3 \leq \pi\}$ ? It's the first step to be a maximum.

For the rest, remember also that $\displaystyle \inf$ (and $\displaystyle \sup$, but it's irrelevant here) can be infinite.
• Aug 2nd 2011, 04:27 AM
lukeheselden
Re: supremum ,infinum ,maximum,minimum
i not sure if i understand that could you state the answers and then it might be more clear to me
• Aug 2nd 2011, 04:48 AM
pece
Re: supremum ,infinum ,maximum,minimum
Well, i'm not gonna give the cooked answers. It's your job to find them.

But I can try to explain clearly.
Let $\displaystyle A = \{q \in \mathbb Q / q^3 \leq \pi\}$. It's your set, right ?

We want to find, if exist, the minimum, the infinum, the maximum and the supremum of $\displaystyle A$.

Let's begin with the supremum (it always exists) : it's by definition $\displaystyle \min \{M \in \mathbb R\cup\{+\infty\} / \forall x \in A, M \geq x\}$ (i.e. the smaller upper bound). You've seen that $\displaystyle \forall x \in A, x \leq \sqrt[3] \pi$ (i.e $\displaystyle \sqrt[3]\pi$ is an upper bound). It's a good start, now you have to show that if $\displaystyle M \in \mathbb R$ is a upper bound, then $\displaystyle M \geq \sqrt[3] \pi$ (by contradiction for example). I let you do that.

Let's continue with the maximum : by definition, a maximum is an upper bound element of the set. So, if a maximum exists, it also is the supremum. So you just have to check if the supremum you found ($\displaystyle \sqrt[3] \pi$) is in $\displaystyle A$. If it is : it's the maximum. If not : there is no maximum.

Now, the infinum (it always exists) : it's by definition $\displaystyle \max \{m \in \mathbb R\cup\{-\infty\} / \forall x \in A, m \leq x\}$. So, what is it here ?

And finally the minimum : it's a lower bound element of the set. Just like the maximum, let's check if the infinum is in the set or not and we have our answer.