Does Tychonov's theorem imply Zorn's lemma I know that Tychonov's theorem, Zorn's lemma, the axiom of choice, the well-ordering theorem and many other results are logically equivalent in the sense that either of them implies all other in ZF. But this is not my question.
I would like to know if there is a direct proof of Zorn's lemma using Tychonov's theorem for a suitably constructed produt of compact spaces.
I am sure that everybody understands the question although it is hard to formalize what I want. But again, this is not the point of this question.
 A: Given a partially ordered set $P$ with the property that every chain has an upper bound, for each chain $C$ let $X_C=\{C\}\cup\{p\in P:\forall c\in C(p> c)\}$. Topologize each $X_C$ with your favorite compact topology that has $C$ isolated (such as $\{\emptyset,\{C\},X_C\}$). By Tychonoff's theorem, $X=\prod_{C} X_C$ is compact.
Note if $X_C\setminus \{C\}$ is empty for some chain $C$, then $C$'s upper bound is a maximal element of $P$, and we're done. So suppose not. For each finite collection of chains $\mathcal F$, $H_{\mathcal F}=\prod_{C} H_{C,\mathcal F}$ defined by $H_{C,\mathcal F}=X_C\setminus\{C\}$ for $C\in\mathcal F$ and $H_{C,\mathcal F}=X_C$ otherwise is a closed subset of $X$ since $\{C\}$ is open in $X_C$. Each $H_{\mathcal F}$ is nonempty: let $f(C)\in X_C\setminus\{C\}\not=\emptyset$ be arbitrary for (importantly, to not require AC, finitely-many) $C\in \mathcal F$, and let $f(C)=C\in X_C$ otherwise. Since $H_{\mathcal F_1}\cap H_{\mathcal F_2}=H_{\mathcal F_1\cup \mathcal F_2}$, the finite intersection of $H_{\mathcal F}$s is also nonempty.
By the FIP characterization of compactness, $\bigcap_{C} H_{C}=\prod_{C}X_C\setminus\{C\}$ is non-empty; choose some $g\in\bigcap_{C} H_C$ and note $g(C)>c$ for all $c\in C$.
We will now embed an order-isomoprhic copy of the class of all ordinals into the set $P$ to obtain our contradiction. First choose $p_0\in P$. Suppose $p_\beta\in P$ has been chosen for all $\beta<\alpha$ such that $p_\gamma<p_\beta$ for all $\gamma<\beta<\alpha$. Then $C_\alpha=\{p_\beta:\beta<\alpha\}$ is a chain, and we may define $p_\alpha=g(C_\alpha)$ where $p_\beta<p_\alpha$ for all $\beta<\alpha$.
Thus we have constructed $\{p_\alpha:\alpha\in\mathbf{On}\}\cong\mathbf{On}$ with $\{p_\alpha:\alpha\in\mathbf{On}\}\subseteq P$, obtaining our contraction.

This proof was obtained essentially by remixing this Wikipedia article with this Math.StackExchange post.
