{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe use the finite intersection property and a maximal family given by Zorn's lemma.\n\n1. Let $X = \\prod_{i \\in I} X_i$ with the [[product topology|04S9HX3GAP]], and let $\\pi_i \\colon X \\to X_i$ be the projections. If $X$ is empty, each open cover of $X$ has the empty subcover, so $X$ is [[compact|0QPHW703PV]]. From now on, $X \\neq \\varnothing$.\n2. A family $\\mathcal{F}$ of subsets of $X$ has the finite intersection property (FIP) when each intersection of finitely many members is nonempty.\n3. By the [[finite intersection property criterion|00BWV5CKCF]] with $K = X$, it suffices to show this. If $\\mathcal{A}$ is a family of [[closed subsets|0G5AV9Q4RR]] of $X$ with FIP, then $\\bigcap \\mathcal{A} \\neq \\varnothing$. Fix such an $\\mathcal{A}$.\n4. Let $P$ be the set of families $\\mathcal{F}$ of subsets of $X$ with $\\mathcal{A} \\subseteq \\mathcal{F}$ and with FIP, ordered by inclusion. It contains $\\mathcal{A}$, so it is nonempty.\n5. Let $\\mathcal{K}$ be a chain in $P$. If $\\mathcal{K}$ is empty, $\\mathcal{A}$ is an upper bound. Otherwise let $\\mathcal{U}$ be the union of the members of $\\mathcal{K}$. It contains $\\mathcal{A}$.\n6. The family $\\mathcal{U}$ has FIP. Finitely many members $D_1, \\ldots, D_n$ of $\\mathcal{U}$ lie in members $\\mathcal{F}_1, \\ldots, \\mathcal{F}_n$ of the chain. The largest of these contains all $D_k$, and it has FIP, so $D_1 \\cap \\cdots \\cap D_n \\neq \\varnothing$.\n7. So each chain of $P$ has an upper bound in $P$. By [[Zorn's lemma|04RXTQA5ZA]], $P$ has a maximal element $\\mathcal{D}$.\n8. Claim: each intersection $E$ of finitely many members of $\\mathcal{D}$ is in $\\mathcal{D}$. Indeed, $\\mathcal{D} \\cup \\{E\\}$ contains $\\mathcal{A}$ and has FIP, since a finite intersection of its members is a finite intersection of members of $\\mathcal{D}$. By maximality, $E \\in \\mathcal{D}$.\n9. Claim: if $S \\subseteq X$ meets each member of $\\mathcal{D}$, then $S \\in \\mathcal{D}$. Take $D_1, \\ldots, D_n \\in \\mathcal{D}$. By step 8, $D_1 \\cap \\cdots \\cap D_n \\in \\mathcal{D}$, so $S$ meets it. Thus $\\mathcal{D} \\cup \\{S\\}$ has FIP, and maximality gives $S \\in \\mathcal{D}$.\n10. Fix $i \\in I$. Let $\\overline{T}$ denote the [[closure|0DRKT1QJR7]] of a set $T$. The family $\\{\\overline{\\pi_i(D)} : D \\in \\mathcal{D}\\}$ consists of closed subsets of $X_i$, by [[this result|0SHC6Z304Z]].\n11. This family has FIP. For $D_1, \\ldots, D_n \\in \\mathcal{D}$, we have $\\pi_i(D_1 \\cap \\cdots \\cap D_n) \\subseteq \\overline{\\pi_i(D_1)} \\cap \\cdots \\cap \\overline{\\pi_i(D_n)}$, and the left side is nonempty by FIP of $\\mathcal{D}$.\n12. The space $X_i$ is compact, so the finite intersection property criterion gives that the set $Z_i = \\bigcap_{D \\in \\mathcal{D}} \\overline{\\pi_i(D)}$ is nonempty.\n13. By the [[axiom of choice|0HTGB13FE2]], choose $x_i \\in Z_i$ for each $i \\in I$, and let $x = (x_i)_{i \\in I} \\in X$.\n14. Let $i \\in I$ and let $U$ be open in $X_i$ with $x_i \\in U$. Claim: $\\pi_i^{-1}(U) \\in \\mathcal{D}$.\n15. Proof of step 14. Let $D \\in \\mathcal{D}$. If $U \\cap \\pi_i(D) = \\varnothing$, then $\\pi_i(D) \\subseteq X_i \\setminus U$, which is closed. Then $\\overline{\\pi_i(D)} \\subseteq X_i \\setminus U$ by [[minimality of the closure|0SHC6Z304Z]], which contradicts $x_i \\in \\overline{\\pi_i(D)}$.\n16. So $U$ meets $\\pi_i(D)$, and hence $\\pi_i^{-1}(U)$ meets $D$. This holds for each $D \\in \\mathcal{D}$, so step 9 gives $\\pi_i^{-1}(U) \\in \\mathcal{D}$.\n17. Let $\\mathcal{S}$ be the [[subbasis|0CHR35YNJQ]] of sets $\\pi_i^{-1}(U)$ with $U$ open in $X_i$. Let $\\mathcal{B}$ be the set of intersections of finitely many members of $\\mathcal{S}$, together with $X$.\n18. Let $\\mathcal{T}$ be the set of all unions of members of $\\mathcal{B}$. It contains $\\varnothing$ and $X$, and it is closed under unions.\n19. The intersection of two unions $\\bigcup_j B_j$ and $\\bigcup_k B'_k$ is $\\bigcup_{j, k} (B_j \\cap B'_k)$, and each $B_j \\cap B'_k$ is in $\\mathcal{B}$. So $\\mathcal{T}$ is a [[topology|0CNQXVBHCX]] that contains $\\mathcal{S}$.\n20. The product topology is the smallest topology that contains $\\mathcal{S}$, so each open set of $X$ is in $\\mathcal{T}$. Thus, if $O$ is open and $x \\in O$, there is $B \\in \\mathcal{B}$ with $x \\in B \\subseteq O$.\n21. Each $B \\in \\mathcal{B}$ with $x \\in B$ lies in $\\mathcal{D}$. For $B = X$, note that $X$ meets each member of $\\mathcal{D}$ and use step 9. Otherwise $B$ is a finite intersection of sets as in step 14, and step 8 applies.\n22. Let $D \\in \\mathcal{D}$. Claim: $x \\in \\overline{D}$. Suppose not. Then $O = X \\setminus \\overline{D}$ is open and contains $x$, so by step 20 there is $B \\in \\mathcal{B}$ with $x \\in B \\subseteq O$.\n23. By step 21, $B \\in \\mathcal{D}$, so $B \\cap D \\neq \\varnothing$ by FIP. But $B \\subseteq O$ and $D \\subseteq \\overline{D}$, so $B \\cap D = \\varnothing$. This contradiction proves the claim.\n24. Let $A \\in \\mathcal{A}$. Then $A \\in \\mathcal{D}$, so $x \\in \\overline{A}$ by step 22. As $A$ is closed, $\\overline{A} = A$, by [[minimality of the closure|0SHC6Z304Z]]. So $x \\in A$.\n25. Thus $x \\in \\bigcap \\mathcal{A}$, which is nonempty. By step 3, $X$ is compact.\n\n## Idea\n\nCompactness means that closed families with the finite intersection property have a common point. We enlarge the given family to a maximal one, which is so large that a point $x$ can be built coordinate by coordinate. In each factor, compactness gives a coordinate $x_i$, and maximality forces each neighbourhood of $x$ into the family, so $x$ lies in the closure of every member.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "54D30"
    ],
    "name": "Proof of Tychonoff's theorem by maximal families with the finite intersection property",
    "proves": "0JTRSJKCFB",
    "sources": [
      {
        "locator": "Proof 2",
        "url": "https://proofwiki.org/wiki/Tychonoff%27s_Theorem"
      }
    ],
    "tag": "15KC9T4MW5",
    "uses": [
      "0QPHW703PV",
      "04S9HX3GAP",
      "0CHR35YNJQ",
      "0CNQXVBHCX",
      "00BWV5CKCF",
      "04RXTQA5ZA",
      "0HTGB13FE2",
      "0DRKT1QJR7",
      "0SHC6Z304Z",
      "0G5AV9Q4RR"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/15KC9T4MW5/",
  "links": [
    {
      "field": "proves",
      "target": "0JTRSJKCFB"
    },
    {
      "field": "uses",
      "target": "0QPHW703PV"
    },
    {
      "field": "uses",
      "target": "04S9HX3GAP"
    },
    {
      "field": "uses",
      "target": "0CHR35YNJQ"
    },
    {
      "field": "uses",
      "target": "0CNQXVBHCX"
    },
    {
      "field": "uses",
      "target": "00BWV5CKCF"
    },
    {
      "field": "uses",
      "target": "04RXTQA5ZA"
    },
    {
      "field": "uses",
      "target": "0HTGB13FE2"
    },
    {
      "field": "uses",
      "target": "0DRKT1QJR7"
    },
    {
      "field": "uses",
      "target": "0SHC6Z304Z"
    },
    {
      "field": "uses",
      "target": "0G5AV9Q4RR"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/15KC9T4MW5.md",
  "msc": [
    "54D30"
  ],
  "name": "Proof of Tychonoff's theorem by maximal families with the finite intersection property",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/15K/15KC9T4MW5.md",
  "status": null,
  "tag": "15KC9T4MW5",
  "uri": "m2://15KC9T4MW5",
  "url": "https://math2.tamnd.com/wiki/proof-of-tychonoffs-theorem-by-maximal-families-with-the-finite-intersection-property/",
  "writing": {
    "tag": "15KC9T4MW5",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}