how to prove supremum and infimum

I started to learn this subject, and I have understood what It's mean, but I cant find out how to find and to prove it. for example :

let $X$ be the set of all rational number in form $m/n$ so $0<m<n$ prove that $\inf X = 0$ and $\sup X =1$, and prove that $\max X$ and $\min X$ does not exist.

how can I solve this? can you please explain it step by step so I would understand

• $\sup X = 0$? 1/2 sits in $X$. – Gibbs Nov 26 '17 at 16:48

Here the supremum case: For any positive integers $m,n$ with $m<n$, then $m/n<1$, so we get immediately $\sup X\leq 1$. Now we claim that $\sup X=1$. Given $\epsilon>0$, by Archimedean property, we can find some positive integer $N$ such that $1/N<\epsilon$, then $(N-1)/N\in X$ and satisfies $(N-1)/N=1-1/N>1-\epsilon$, so $\sup X>1-\epsilon$. Since this is true for all $\epsilon>0$, then $\sup X\geq 1$, so we get $\sup X=1$.

Similar argument using Archimedian property will show that $\inf X=0$.

So your set is $$X = \bigg\{\frac{n}{m}: n, m \in \mathbb{N} \mbox{ and } 0 < m < n\bigg\}.$$ Claim: $\inf X = 0$ and $\sup X = 1$. You can easily see that both of them do not sit in $X$ as $m$ cannot be $0$ and $m$ cannot be equal to $n$. So in case we prove our claim, we can say that $0$ and $1$ are not minimum and maximum of $X$. This is because maxima and minima are required to be in your set in general. To check that $0$ is the infimum you can check two properties: that $0$ is smaller than every element of your set, and that if you pick $\varepsilon > 0$ you can always find an element of your set between $0$ and $0+\varepsilon$.

Clearly $0 < \frac{m}{n}$ as $m,n > 0$. So let us find $\frac{m}{n} < \varepsilon$. Let us fix $l \in \mathbb{N}, l > 1$ large enough so that $l\varepsilon > 1$. Such an $l$ exists by the Archimedean property of the real numbers. So $\frac{1}{l} < \varepsilon$. This proves your first claim, $\inf X = 0$.

You can do a similar proof for $\sup X = 1$. It is enough to prove that $1$ is greater than all the elements in $X$, and that once you fix $\varepsilon > 0$ you can always find an element of $X$ between $1-\varepsilon$ and $1$.

First, we show the infimum and supernumerary are as you say. Clearly $0$ is a lower bound, and $1$ an upper bound, but we must show they are, respectively, the greatest lower bound and least upper bound. I claim that (i) if $a > 0$ is a rational number, then there exists $x \in X$ such that $x < a$, and (ii) if $b < 1$ is a rational number, then there exists $y \in X$ such that $y > b$.

For (i), let $a = m / n$, where $0 < m < n$. We wanna find a number between $0$ and $a$, so we can make the easy choice and pick the midpoint between them, namely $x = \frac{m}{2n}$. Thus if $a$ was

For (ii), let $b = p / q, 0 < p < q$, and then let $y = \frac{2p + 1}{2 q}$. Again, we choose a point between $b$ and $1$ to show $b$ wasn't an upper bound.

So now we've determined that $\inf X = 0, \sup X = 1$. But neither $0$ nor $1$ is contained in $X$, so they can't be the minimum or maximum, respectively.