Zorn's Lemma states that if every chain $C$ in a partially ordered set $X$ has an upper bound then there is at least one maximal element in $X$.

Let $n\in \mathbb{N}$, $C_n \subset \mathbb{N}$ be the set containing the natural numbers $\{1,\dots,n\}$. Then if $k\le n$, $C_k\subset C_n$, so $C_n$ is a chain. Also, $C_n$ has an upper bound, such as $n+1$. But what is the maximal element of $\mathbb{N}$?

There's something I don't get in Zorn's Lemma statement.

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    $\begingroup$ Well, your $C_n$ does not include every chain! $\endgroup$ Sep 14 '17 at 0:39
  • $\begingroup$ @I.Padilla Why does it need to include every chain? $\endgroup$
    – sequence
    Sep 14 '17 at 0:52
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    $\begingroup$ @sequence Because Zorn's Lemma only applies if every chain has an upper bound. All you've shown is that the specific chains $C_n$ have upper bounds; so what? $\endgroup$ Sep 14 '17 at 1:03
  • $\begingroup$ Um... because that's a condition for Zorn's Lemma--- that every chain has an upper bound. If C has no upper bound then it can't have a maximum element. Zorn's Lemma does not apply to $\mathbb N$ because not every chain has an upper bound. $\endgroup$
    – fleablood
    Sep 14 '17 at 1:04
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    $\begingroup$ $C_n$ satisfies Zorn's Lemma and $C_n$ does have a maximal element. But $C_n$ is not the same thing as $\mathbb N$. For Zorn's lemma to apply to $\mathbb N$ then every chain in $\mathbb N$ must be bounded above. That is not true for $\mathbb N$. $C_e=\{2n|n \in \mathbb N\} = \{$even numbers\}$ is a chain that is not bounded above. So Zorn's lemma does not apply to N. You listed a bunch of chains that were bounded above--- but they were not all of the chains in $\mathbb N$. So they don't matter. $\endgroup$
    – fleablood
    Sep 14 '17 at 1:30

You just take some examples of chains, but there are many others, for example the multiples of any positive integer is a chain. The hypothesis of the lemma is that every chain must have a upper bound, but $\mathbb{N}$ is a total ordered set, then every subset of $\mathbb{N}$ is a chain, in special, $\mathbb{N}$ itself is a chain.

The Zorn's lemma is quite useless in a total ordered set, because, if this set satisfies the hypothesis, the upperbound of the set itself (which is a chain) is the maximal element.


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