# Simple and intuitive example for Zorns Lemma

Do you know any example that demonstrated Zorn's Lemma simple and intuitive? I know some applications of it, but in these proof I find the application of Zorn Lemma not very intuitive.

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Many of the most straightforward applications of Zorn’s lemma can be seen as slightly disguised applications of the Teichmüller-Tukey lemma. I doubt that you can make either very intuitive, but the special case of sets of finite character seems a little easier to recognize and get a feel for than the general case. – Brian M. Scott Dec 5 '12 at 23:33

Zorn's lemma is not intuitive. It only becomes intuitive when you get comfortable with it and take it for granted. The problem is that Zorn's lemma is not counterintuitive either. It's just is.

The idea is that if every chain has an upper bound, then there is a maximal element. I think that the most intuitive usage is the proof that every chain can be extended to a maximal chain.

Let $(P,\leq)$ be a partially ordered set and $C$ a chain in $P$. Let $T=\{D\subseteq P\mid C\subseteq D\text{ is a chain}\}$. Then $(T,\subseteq)$ has the Zorn property, because a chain in $T$ is an increasing collection of chains. The $\subseteq$-increasing union of chains is a chain as well, so there is an upper bound. By Zorn there is a maximal element, and it is a maximal chain by definition.

If you search on this site "Zorn's lemma" you can find more than a handful examples explaining slightly more in details several discussions and other applications of Zorn's lemma. Here is a quick list from what I found:

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It is not intuitive. It's not without reason that this joke is well known:

"The Axiom of Choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?" — Jerry Bona

In particular note that the upper bound created during a proof using Zorn's Lemma does not have to be a maximal element. As K. Conrad remarks in Example 2.3, the ideals $\{6\mathbb Z, 12\mathbb Z, 24\mathbb Z\}$ in $\mathbb Z$ have $6\mathbb Z$ as an upper bound (with the partial ordering just being set inclusion), but $6\mathbb Z$ is clearly not maximal among proper ideals in $\mathbb Z$.