Ytukyg

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

Here the supremum case: For any positive integers $m,n$ with $m<n$, then $m/n<1$, so we get immediately $sup Xleq 1$. 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 $(N-1)/Nin X$ and satisfies $(N-1)/N=1-1/N>1-epsilon$, so $sup X>1-epsilon$. Since this is true for all $epsilon>0$, then $sup Xgeq 1$, so we get $sup X=1$.

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

So your set is
$$X = bigg{frac{n}{m}: n, m in mathbb{N} mbox{ and } 0 < m < nbigg}.$$
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 $lvarepsilon > 1$. 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.