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  "body": "\n## Proof\n\nZorn's lemma implies the ultrafilter lemma, which in turn implies [[Tychonoff's theorem|0JTRSJKCFB]] (together with the axiom of choice). However, it is also possible to prove Tychonoff's theorem directly from Zorn's lemma as follows (by using an ultrafilter implicitly).\n\nLet $X_i$ be a family of [[compact spaces|0QPHW703PV]], not necessarily Hausdorff. We shall show that a family $F$ of closed subsets of $X := \\prod_i X_i$ with the [[finite intersection property|00BWV5CKCF]] has nonempty intersection. Let $P$ be the collection of all the families of subsets of $X$ with the finite intersection property (not necessarily closed subsets). We let $P$ be ordered by set inclusion. If $C \\subset P$ is a chain, then clearly the union of $C$ has the finite intersection property; so, the union is in $P$. Thus, by Zorn's lemma, there is a maximal element $M$ in $P$ containing $F$. We shall show that the intersection of all the closed sets in $M$ has nonempty intersection; *a fortiori*, $F$ has nonempty intersection.\n\nFor each finite subset $M' \\subset M$, since $\\cap M'$ is nonempty, $\\cap p_i(M')$ is nonempty for the projection $p_i : X \\to X_i.$ Thus, for each $i$, the set\n\n$$N_i := \\left \\{ \\overline{p_i(A)} \\mid A \\in M \\right \\}$$\n\nhas nonempty intersection since it has the finite intersection property and $X_i$ is compact. We note\n\n1. If $M' \\subset M$ is a finite subset, $\\cap M'$ is in $M$.\n1. For an open set $U_i$ intersecting $\\cap N_i$, we have $p_i^{-1}(U_i)$ is in $M$.\n\nIndeed, (1) holds since $M = M \\cup \\{ \\cap M' \\}$ by maximality, as the set on the right has the finite intersection property. Similarly, for a finite subset $M' \\subset M$, since $A = \\cap M'$ is in $M$, $U_i \\cap p_i(A) \\ne \\emptyset$ or $p_i^{-1}(U_i) \\cap A \\ne \\emptyset$. Thus, the set $M \\cup \\{ p_i^{-1}(U_i) \\}$ has the finite intersection property and (2) follows by the maximality of $M$.\n\nNow, by the axiom of choice, we can find an element $m$ in $\\prod_i (\\cap N_i)$. We claim this $m$ is in each closed set $A$ in $M$. Since $A$ is closed, it is enough to show $m$ is in the closure of $A$; i.e., each basic neighborhood of $m$ intersects $A$. A basic neighborhood of $m$ has the form $\\cap_{j=1}^r p_{i_j}^{-1}(U_{i_j})$. But by the property (2) above, we have $\\left \\{ A, p_{i_j}^{-1}(U_{i_j}) \\mid j = 1, \\dots, r \\right \\} \\subset M$, which gives the claim by the finite intersection property. $\\square$\n",
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        "locator": "Zorn's lemma, section Example applications \u003e A proof of Tychonoff's theorem",
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