No largest element of Dedekind cuts

Question

I'm practicing proving that various sets are (or aren't) Dedekind cuts. I am struggling with showing the property that there is no largest element. For example, for a simple Dedekind cut:

$A=\{{a\in\mathbb{Q}:a^2<2}$ or ${a<0}\}$

what is a good general method for showing that: if ${a\in A}$, there is a ${b}$ in ${A}$ with ${a<b}$

if $a\le0$, take $b=1$; if $a>0$, take $b=\dfrac{2a+2}{a+2}$; cf. this question — J. W. Tanner, Feb 13 '20 at 23:27
Thank you. I still don't understand how you know to define it that way. — mayalarson, Feb 14 '20 at 03:43
Out of all the properties of Dedekind cuts the one you are having trouble with is the most difficult to deal. And that marks the beginning of analytical argument. Other properties of Dedekind cuts are verified using what you can call an algebraical argument. — Paramanand Singh, Feb 14 '20 at 04:53

Paramanand Singh · Answer 1 · 2020-02-14T04:51:23.063

Here is how one would approach this problem if one has no prior experience with such problems.

You are given $a\in A$ and you wish to find $b\in A$ with $a<b$. There are two cases to consider: first $a\leq 0$ and then we can simply take $b=1$.

The real challenge is to deal with case $a>0$. The problem can now be stated as

Let $a$ be a positive rational number with $a^2<2$. Prove that there is another positive rational number $b>a$ such that $b^2<2$.

Thus we want to have $b$ such that $a^2<b^2<2$. Now let us put $b=a+h$ with $h>0$ and we need to find $h$ which ensures $b^2=(a+h)^2<2$. Thus we need $$a^2+2ah+h^2<2$$ or $$h(2a+h)<2-a^2$$ Let's now observe that the left hand side decreases if $h$ decreases and hence we need some sort of small value of $h$ here. Let's arbitrarily restrict $h<1$ (you can choose $h<2$ or whatever you wish). Then we have $$h(2a+h)<h(2a+1)$$ and if we further ensure that $h(2a+1)<2-a^2$ then our job is done. Thus we need two conditions on $h$ namely $h<1$ and $$h<\frac{2-a^2}{2a+1}$$ and combining this we get $$0<h<\min\left(1,\frac{2-a^2}{2a+1}\right) $$ and then $b=a+h$ is our required element of $A$.

Let's check how this works out in practice. Let $a=1.4$ so that $a^2=1.96<2$. Now we have $$\frac{2-a^2}{2a+1}=\frac{0.04}{3.8}<1$$ and we need to find a positive $h$ less than above value. Clearly one such value is $$h=\frac{0.04}{4}=0.01$$ and hence $b=a+h=1.41$ works fine. Check that $1.41^2<2$. You can try another example with $a=1.41$. The key is that the entire argument in last paragraph is symbolic and deals with any positive value of $a$ with $a^2<2$. And thus given any positive $a\in A$ we are able to find another $b\in A$ with $b>a$.

You should try the slightly harder problem using the same approach :

Let $a$ be a positive rational number such that $a^3<3a+8$. Prove that there is another positive rational number $b$ with $b>a$ and $b^3<3b+8$ so that the set $$B=\{x\mid x\in\mathbb {Q}, x>0,x^3<3x+8\}$$ has no greatest member.

No largest element of Dedekind cuts

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