Applying Zorn's Lemma to natural numbers

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.

• Well, your $C_n$ does not include every chain! – I.Padilla Sep 14 '17 at 0:39
• @I.Padilla Why does it need to include every chain? – sequence Sep 14 '17 at 0:52
• @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? – Noah Schweber Sep 14 '17 at 1:03
• 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. – fleablood Sep 14 '17 at 1:04
• $C_n$ 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$. So they don't matter. – fleablood Sep 14 '17 at 1:30 1 Answer You just take some examples of chains, but there are many others, for example the multiples of any positive integer is a chain. The hypothesis of the lemma is that every chain must have a upper bound, but$\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.