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Is there anything wrong with the following proof?

Theorem. Every non-empty subset $B \subset\mathbb{N}$ has a least member.

Proof. Assume not. Then, of necessity, we'd have to have $B=\varnothing$, for if $B$ contained even one $b\in\mathbb{N}$, then that $b$ would satisfy $b=min(B). $ Contradiction, end proof.

I get the sense there is something wrong here, but I can't seem to define exactly what. Also, is there a context where one can simply take this principle as an axiom, and not have to prove it? After all, it is extremely intuitive. Thanks.

Andrés E. Caicedo
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Coffee_Table
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5 Answers5

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You should be more detailed in what you claim. Your proof is wrong, and it is very hard to understand why you obtain said conclusions. My point is:

"Then, of necessity..." Why?

"...if $B$ contained even one $b∈\Bbb N$, then that $b$ would satisfy $b=\min(B)$? Why?

A correct proof would go as follows:

P Let $B\subseteq \Bbb N$ be nonempty. We prove by induction that $B$ has a least element. Assume by contradiction that $B$ has no least element. Let $J$ be the set of elements that are not in $B$. Since $0$ is a lower bound of $\Bbb N$, $0\notin B$ (else it would be a least element) so $0\in J$. We proceed to prove the induction step $0,1,\dots,n\in J\implies n+1\in J$. Indeed, suppose that $0,1,\dots,n\in J$. Then $n+1$ cannot be in $B$ since then it would be a lower bound of $B$, and since $0,1,\dots,n\notin B$, it would be a least element. It follows $n+1\in J$. By induction, $J=\Bbb N$ so $B=\varnothing$ which is impossible.

As Pete is saying, WOP is equivalent to PMI.

PROP Suppose every nonempty subset of $\Bbb N$ has a least element. Let $B$ be a subset of $\Bbb N$ with the following properties

$(1)$ $0\in B$.

$(2)$ $n\in B\implies n+1\in B$

We show that $B=\Bbb N$.

P Let $B$ be as above. Consider the set of $\Bbb N\setminus B$, and assume by contradiction it is not empty. By the WOP, it has a least element, call it $a$. Since $0\in B$, this element must be of the form $a=n+1$ for some $n\in \Bbb N$. Since $n+1$ is the first element that is not in $B$, $n$ is an element of $B$. But then $n+1\in B$, which is absurd. It follows that $\Bbb N\setminus B$ must be empty, so $B=\Bbb N$, as claimed.

yellowgrass
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Pedro
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What do you mean by "if $B$ contained even one $b \in \mathbb{N}$?" Your argument works if $B$ contains exactly one $b$. It doesn't seem to work otherwise: e.g. the subset $\{1,3\}$ contains "even one element $3$", but $3$ is not the least element.

Also, yes: you can take WOP as part of an axiomatic description of the natural numbers $\mathbb{N}$. You should know that -- for instance, given the other four Peano axioms -- it is equivalent to the Principle of Mathematical Induction.

Pete L. Clark
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    Indeed, now I see my error. My "proof" misses the essential condition of possibly having more than one element in the set! For in that case, what I've said shows absolutely nothing. – Coffee_Table Apr 12 '13 at 01:42
  • However, why does the general result not follow from the fact that natural numbers obey inequalities? That is, we can always arrange them in some sequence $n_0<n_1<...$, so that we don't need induction? I'm missing something here... – Coffee_Table Apr 12 '13 at 01:43
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    @Coffee_Table Well, but how do you pick $n_0$?. You're assuming there exists a least element (namely, this $n_0$), when you should be proving it. =) – Pedro Apr 12 '13 at 01:51
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    @PeterTamaroff I see. Like Andy Brandi said as well, the root of all of my errors here is that I already assume that I can work with this entity called $min(b)$; but that reasoning is the height of circularity! Subtle, but I see fallacy now in all of this. Thanks. – Coffee_Table Apr 12 '13 at 01:55
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    How do you prove the Principle of Mathematical Induction from WOP and the other 4 axioms? – Vladimir Reshetnikov Jul 29 '21 at 19:42
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    https://en.wikipedia.org/wiki/Mathematical_induction#:~:text=it%20is%20mistakenly%20printed%20in%20several%20books%5B22%5D%20and%20sources%20that%20the%20well-ordering%20principle%20is%20equivalent%20to%20the%20induction%20axiom%2E – Vladimir Reshetnikov Jul 29 '21 at 19:50
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Here is another proof from the book Topology by Munkres:

enter image description here

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    is it obvious that the smallest element of intersection of D and {1, ..., n} is the smallest element of D as well? – mathslover Feb 27 '21 at 17:36
  • @mathslover The smallest element $k$ is in ${1,…,n}$, which is equivalent to $k\leq n$. We know that $k\in D$, but suppose $k$ is not the smallest element of $D$. Then $\exists d\in D$ such that $d<k$ and thus $d\leq n$ and thus $d\in D\cap{1,…,n}=A$. But $k$ is the smallest element of $A$, so $k\leq d$, which is a contradiction. – Ryan Bell Aug 09 '24 at 01:48
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Assume not. Then, of necessity, we'd have to have $B=\varnothing$, for if $B$ contained even one $b\in\mathbb{N}$, then that $b$ would satisfy $b=\min(B). $

This only works if $B$ contains exactly one element. For any other $B$, you have to assume that $\min$ is well-defined, i.e. every subset of $\mathbb{N}$ has a least member, so your reasoning is circular.

Andy Brandi
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Would this work as a proof of the well ordering principle alone? Perhaps not at all the "if POI is true then WOP" part...

Assume that $S$ is a non empty subset of $\mathbb{N}$ and that it is has no least element. Then, there must exist an $m$ such that $m$ is a natural number and an element of $S$ as otherwise $S$ would be empty. Then: $$inf(\mathbb{N}) \le inf(S) \le m$$ since if $S$ had included a lesser element than $inf(\mathbb{N})$ that element wouldn't be in $\mathbb{N}$ and it is bounded above by $m$ because we know $m \in S$ we don't need anything greater than m to be the least infimum of $S$ as there is always $m$. So $inf(S)$ is bounded in a finite region which suggests that there exists a least element in $S$, this is a contradiction.

AHusain
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guest4104
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