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I have been reviewing my intro to analysis course and stumbled on a rather easy proof. The proof is standard, suppose $\mathbb{N}$ is bounded above and reach a contradiction using the Completeness Axiom. My question is why do we even need to do that? Consider the following proof:

To show $\mathbb{N}$ is not bounded above is the same as showing that given $n_1\in\mathbb{N}$, there exits $n_2$ s.t. $n_2>n_1$ and $n_2\in\mathbb{N}$. Letting $n_2=n_1+1$ completes the proof. Why do we even need to use the completeness axiom if we already know $\mathbb{N}\subset \mathbb{R}$.

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    $\begingroup$ The assertion is intuitively reasonable, there is no real number that is greater than every integer. However, there are important "reals-like" structures in which $\mathbb{N}$ is indeed bounded above. $\endgroup$ – André Nicolas Apr 18 '16 at 1:20
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    $\begingroup$ $\mathbb{N}$ is bounded from above in the set of countable ordinals and your proof still holds in that case so there is something wrong with your proof. $\endgroup$ – Q the Platypus Apr 18 '16 at 1:20
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    $\begingroup$ @user1952009, I don't think its ridiculous. How do you know that that the only model of the complete ordered field axioms is $\mathbb{R}$? Maybe it has other, non-isomorphic models. To check that this isn't the case, you need to prove that every complete ordered field is Archimedean. $\endgroup$ – goblin Apr 18 '16 at 5:28
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    $\begingroup$ @user1952009, if I were trying to communicate with the OP, I'd be using more more-familiar (but also more long-winded) words and phrases. I'm using those "big complicated words" to talk to you, not the OP. In any event, my point remains, which is that to check that the relevant axioms pin down our intuitions about $\mathbb{R}$, you have to prove these sorts of things. What's so ridiculous about that? $\endgroup$ – goblin Apr 18 '16 at 5:52
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    $\begingroup$ I think you missed the whole point of "bounded above". To show that $\mathbb{N}$ is not bounded above you must show that if $K$ is any real number then there is an $n \in \mathbb{N}$ such that $K < n$. What you are trying to do is to find a natural number greater than a given natural number. This has nothing to do with "bounded above", rather it has something to do with the infinite cardinality of set $\mathbb{N}$. $\endgroup$ – Paramanand Singh Apr 18 '16 at 7:23
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Saying that $\Bbb N$ is not bounded above in $\Bbb R$ is not the same as saying that for all $x\in \Bbb N$ there exists $y\in \Bbb N$ such that $y>x$. In fact, there are several subsets of $\Bbb R$ which satisfy this property and are bounded above, namely the interval $[0,1)$, which has $1$ as an upper bound. What you must prove is that for all $x\in\Bbb R$ there exists $y\in \Bbb N$ such that $y>x$, which is different.

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    $\begingroup$ I didn't realize the distinction between $\mathbb{N}$ is not bounded above in $\mathbb{R}$ and $\mathbb{N}$ is not bounded above in itself. The latter will be trivial, right? $\endgroup$ – Kun Apr 18 '16 at 1:14
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    $\begingroup$ @Kun The latter is trivial and your proof is valid, but they are not the same. For example, $(0, 1)$ is not bounded above in itself. Try to find a biggest element of $(0, 1)$. Can you do it? Are you sure? (If you find a biggest element, tell me and I will find a bigger one.) Even though $(0, 1)$ has no biggest element, $1$ is still a upper bound of the set. $\endgroup$ – Noble Mushtak Apr 18 '16 at 1:17
  • $\begingroup$ @Kun Per se, the fact that the ordering of $\Bbb N$ has no maximum should follow easily by its set-theoretic definition. If i recall correctly my last ZFC exam, proving the good definition of the ordering of $\Bbb N$ from scratch was a bit tedious. $\endgroup$ – user228113 Apr 18 '16 at 1:21
  • $\begingroup$ Wait...Assuming $0 \notin \Bbb{N}$, can't you just define $<$ as $\{(a, b) \in \Bbb{N}^2 \mid (\exists c \in \Bbb{N})(a+c=b)\}$ and then prove the relation is irreflexive, transitive, and asymmetric? That seems pretty easy. Then, to prove it has no maximum, you just prove the Well-Ordering Theorem and go from there. $\endgroup$ – Noble Mushtak Apr 18 '16 at 1:24
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    $\begingroup$ @NobleMushtak Perhaps, but then you need to have already proven some things on the sum, namely the fact that $$a+c=a\implies c=0$$ Discussing it here in detail would be long, I fear. $\endgroup$ – user228113 Apr 18 '16 at 1:29
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Maybe a picture will help. $\;\!\;\!$

enter image description here

So the purpose of the "lengthy proof" is to show that complete ordered fields are never too lengthy. Now that's irony!

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Your proof shows that there is no biggest element of $\Bbb{N}$. However, just because there is no biggest element of $\Bbb{N}$ does not mean there is no upper bound.

Take $(0, 1)$, for example. If you have some biggest element $m \in (0, 1)$, then we know $m < 1$, so $1-m > 0$ and thus $m+\frac{1-m}{2}=\frac{1+m}{2} > m$. Clearly, since $m < 1$, $1+m < 2$, so $\frac{1+m}{2} < 1$, meaing $\frac{1+m}{2} \in (0, 1)$. Thus, we have found a bigger element than the biggest element of $(0, 1)$, so there is no biggest element of $(0, 1)$.

However, we can not deduce from the fact that $(0, 1)$ has no biggest element that it is not bounded above. Clearly, $1 \in \Bbb{R}$ is an upper bound of $(0, 1)$ even though $(0, 1)$ has no biggest element.

Thus, even though it seems obvious that $\Bbb{N}$ has no upper bound, we need to use the Completeness Axiom in order to prove it because that tells us about the real numbers themselves. In fact, there are some models of the real numbers like the hyperreals where $\Bbb{N}$ is actually bounded above because the Completeness Axiom does not hold. However, your proof that $\Bbb{N}$ has no biggest element would still hold in the hyperreals, but that doesn't mean it has no upper bound.

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