# Why does Dedekind completness implies that $A$ has an infimum and a supremum?

So I have the sets $$A:=\left\{\frac{1}{n}: n \in \mathbb{N}\right\}$$ and $$B: = \{ x \in \mathbb{Q}:x^3\gt 2 \}$$.

The exercise asks me to explain why $$\inf A$$, $$\sup A$$ and $$\inf B$$ exists.

I can see that $$A=(0,1] \in \mathbb{R}$$. I can see that $$A$$ has a $$\inf A$$ and $$\sup A$$. This is probably not a sufficient argument but this is all I can do right now. For $$B$$ I'm a bit lost since the rational numbers do not have a sup and inf.

Kind regards

I have been doing some more thinking. This is what I achieved :

I want to prove that $$\inf A$$ exists. It is sufficient to show a lower bound exists. By looking at the elements in $$a_n = \frac{1}{n}\in A$$, I can see that $$0 \lt a_n =\frac{1}{n}$$. So my guess for a lower bound is $$0$$. Now I want to prove that this is indeed a lower bound. If I can prove this, then I have proven the existence of $$\inf A$$. So I will like to show that $$a_n =\frac{1}{n}$$ can come arbitrarily close to $$0$$. So now I begin: $$|a_n - 0|\lt\varepsilon \implies\frac{1}{n}\lt\varepsilon\implies n\gt\frac{1}{\varepsilon}$$. By the archimedean property, a natural number $$n$$ exists. Conclusion is that a lower bound for $$A$$ exists. By the Definition of Dedekind completeness, a least lower bond ($$\inf A$$) also exists. $$\square$$. One thing which I'm not sure about is if I show that $$|a_n-0| < \varepsilon$$ holds for all $$\varepsilon > 0$$, then I have shown that $$1/n = 0$$ (the distance between is $$0$$). But this is clearly not the case since $$1/n >0$$. I do not know how do make an argument for that.

• Be careful! $A \neq (0,1]$ - After all, for which $n$ is $1/n = 1/\pi$? It is true, however, that $A \subseteq (0,1]$. – HallaSurvivor Apr 18 at 0:45
• Even $A\subset(0,1]\cap\mathbb Q$, it's even obvious that $A\cap\left[\frac{1}{2},1\right]=\left\{\frac{1}{2},1\right\}$. Also,$A\color{red}{\subset}\mathbb R$, but $A\color{red}{\in}\mathcal P(\mathbb R)$! – Croissant Apr 18 at 11:10
• @HallaSurvivor Thanks for the notice. I can see this now. – Xenusi Apr 18 at 11:15
• Also, it isn't sufficient to prove $0$ is just a lower bound, it's the greatest lower bound. – Croissant Apr 18 at 11:19
• You mean $A$ is incomplete? Yes, I can see that. But why is that a problem? Btw. I only have 1/2 page of lecture notes about Dedekind completness, so my understand could be (probably is) full of holes :-). – Xenusi Apr 18 at 11:27

Hint:

Recall the definition of completeness:

Any set $$A \subseteq \mathbb{R}$$ with an upper bound (resp. lower bound) has a supremum (resp. infimum)

So to show $$\inf A$$ exists, it suffices to provide a lower bound. As you have noticed, $$0 < \frac{1}{n}$$ for every $$n$$, and so $$0$$ is the required lower bound for $$A$$.

Can you use this same kind of argument to show $$\sup A$$ exists? What do you need to show to guarantee a supremum exists?

For $$B$$, notice you are only asked to find an infimum - this is because the supremum doesn't exist. Why? Because there is no upper bound! I can always find bigger and bigger choices for $$x$$ making $$x^3 > 2$$. After all, $$10^3 > 2$$, $$100^3 > 2$$, $$1000^3 > 2$$, and so on.

But why must $$B$$ have an infimum? If you can show it has a lower bound, then you're done. Can you find a number $$k$$ so that $$k < x$$ for every $$x \in B$$?

Good luck!

I hope this helps ^_^

As to your first remark, I want to point out that A is very obviously not $$(0, 1]$$, although it is true that $$A \subseteq (0, 1]$$. You are right in thinking that $$\inf A = 0$$. Clearly all $$a \in A$$ are greater than $$0$$. You just need to show that the elements of $$A$$ can be arbitrarily close to $$0$$, which made precise is to say for any $$\varepsilon \in \mathbb R_{>0}$$, we can find some $$a \in A$$ so that $$a < 0 + \varepsilon$$, I think this will be easy (Hint: are there integers larger than $$1/\varepsilon$$?).

To your question about $$B$$, it might be easier to translate the set by observing that a rational number satisfies $$x^3 > 2$$ if and only if it satiesfies $$x > 2^{1/3}$$. Hence we really have $$B = \{x \in \mathbb Q : x > 2^{1/3}\}$$. It is now very easy to see what its infimum is, and the proof is essentially the same as for $$A$$.

Edit: The core idea of your proof is correct, that there exists $$n \in \mathbb N$$ greater than $$1/\varepsilon$$. But the line of logic is flawed. Let me elaborate:

1. I'm not sure what exactly the Dedekind completeness is defined as (it could be one of many equivalent properties depending how your class/material is teaching), but I will assume (based on the way you used it) that it means "a set in $$\mathbb R$$ that is bounded below has an infimum". In that case you don't need to compute that $$\inf A = 0$$, you only need to show that $$A$$ is bounded below. For instance it is bounded below by $$-1$$ (every element of $$A$$ is greater than or equal to $$-1$$), and that's good enough to invoke the completeness property -- an infimum exists.

2. Remember that the definition of $$\inf A$$ is $$\max \{x \in \mathbb R: x \leq a,\ \forall a \in A\}$$. That means you have to show first that $$0$$ is less than or equal to all elements in $$A$$, and that any number $$\varepsilon$$ larger than $$0$$ must not be less than or equal to all of $$A$$, i.e. you can find some $$a\in A$$ that is smaller than $$\varepsilon$$. You certainly don't need a specifict $$a \in A$$ to satisfy this for all of $$\varepsilon$$. Instead, for every $$\varepsilon$$ you need to find such an $$a$$, which you did correctly.

3. $$|a_n-0|< \varepsilon$$ is not what you want, this is good for showing that the sequence $$\{a_n\}$$ converges to $$0$$, but for the $$\inf$$ you need specifically $$a_n < \varepsilon$$, see point 2.

4. The line of logic that $$a_n < \varepsilon \implies n > 1/\varepsilon$$ is not the correct to argue, and I see this in many beginners, because you work backwards from the claim to the properties that you need. But this is the wrong way to present your argument, what you really need to say is that $$n > 1/ \varepsilon \implies a_n < \varepsilon$$, that something you know is true implies something you want to prove is true. Certainly in this case the implication is bidirectional (it's true as an "if and only if"), but in many proofs it will not be, and you need to be careful with that.

• I have wrote an answer in my post based on what you have said. Could you please let me know if I have understand you correctly? It's too long to post as a comment to your answer. – Xenusi Apr 18 at 11:00
• @Xenusi Sure, I've made an edit to comment on your proof, hope it helps. – poopist Apr 18 at 14:30
• Okay, thanks! I'm going to read it! – Xenusi Apr 18 at 14:34