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I am trying to understand the proof of archimedean priniciple stated on wiki proof here.(https://proofwiki.org/wiki/Archimedean_Principle). I am having trouble understanding the last part of the proof where it proves there is a supremum $$s = sup(S)$$ and then proceeds with 'Now consider a number $s - 1$' and then eventually proving that $m + 1 > x$. From what i understood, for example if S = {1, 2, 3, 4, 5, 6, 7, 8 , 10} and x=10,then could the supremum of S be 10.9999999999 or 11 ? . If $ s = 11 $ then the last part of the proof does not make sense, i.e if $m=10, s = 11$ then $m > s - 1$ which evalutes to $10 > 10$ which is false. So am i correct in saying that s is 10.99999999 ?

José Carlos Santos
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jnxd
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    If $S={1,2,3,4,5,6,7,8,9,10}$, then $s=\sup S=10$. – José Carlos Santos Oct 09 '21 at 18:38
  • But doesnt the proof states that a supremum could be any real number outside the set S ? – jnxd Oct 09 '21 at 18:58
  • The supremum of a set, when it exists, can only be one number. Where does it say that it could be any real number outside the set $S$? – José Carlos Santos Oct 09 '21 at 19:00
  • It seems as if the concepts of upper bound and least upper bound are being confused with each other. – user2661923 Oct 09 '21 at 19:07
  • It says in the proof like this "Thus by the Continuum Property of R, S has a supremum in R.

    Let s=sup(S)." So does it mean the supremum lies with in the set S or outside of S in R ?

     – jnxd Oct 09 '21 at 19:09
  • @user2661923 Not sure , please enlighten me, the least upper bound in the above example is either 10.99999 or 11 rite ? The next least upper bound if there is such a thing is 11.99999999 or 12 – jnxd Oct 09 '21 at 19:12
  • If $S = {a \in \Bbb{N} ~: ~a\leq 10}$, then the following elements from $\Bbb{N}$ are all upper bounds to $S$ : ${10,11,12,13,14,15,\cdots}$, while the following subset of $\Bbb{R}$ would be the complete set of all upper bounds to $S ~: ~{a \in \Bbb{R} ~: ~a \geq 10}.$ The least upper bound to $S$ is $(10)$. – user2661923 Oct 09 '21 at 19:17
  • For some sets $S$, $\sup(S)\in S$, whereas for others we have $\sup(S)\notin S$. It depends. In the case when $S={1,2,3,4,5,6,7,8,9,10}$, $\sup(S)=10\in S$. – José Carlos Santos Oct 09 '21 at 20:11
  • Thanks all. In the example above m would be 10 , correct? . That would evaluate $m > s-1$ to $10 > (10 - 1)$ which is true for this specific example. – jnxd Oct 10 '21 at 00:30

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Just to elaborate on the proof further in case that it could help.

By the continuum property of $\mathbb{R}$ they are invoking that

If a $S$ is a non-empty subset of the reals $\mathbb{R}$ and bounded above, then $S$ has a least upper bound (Supremum).

The reason the proof takes a look at $s-1$, where $s=\sup(S)$. Is to show that this is not an upper bound, for if $s-1$ is an upper bound for $S$, then $s-1<s$ contradicting that $s$ is the least upper bound of $S$.

Therefore negating what is means for an element to be an upper bound, that means that since $s-1$ is not an upper bound, there must be at least one element $m$ in $S$ such that $s-1<m$.

Since $m$ is an element of $S$, then $m$ is a natural number. Finally $s-1<m\iff s<m+1$, that means $m+1\notin S$ yet it's a natural number so $m+1>x$.

Now if the logic above was not what was bothering you about the proof but rather some misconception about the continuum property (axiom of completeness), I suggest reading up on it since it's a really neat property of the real numbers.

I don't have enough reputation to comment but I saw your question on whether supremums of sets need or need not be in the sets.

Take the set $$A=\{x\in\mathbb{R}: 1\leq x<2\}: n\in\mathbb{N}\}$$

Here we have $\sup(A) = 2$ (this can be proven) yet $2\notin A$

While for $$B=\{x\in\mathbb{R}: 1\leq x\leq 3\}: n\in\mathbb{N}\}$$

we have $\sup B = 3$ and $3\in B$.

Gabríel Snær Völuson
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