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  "body": "\n## Definition\n\nLet $X$ be a [[topological space|0CQHFPGJ45]].\n$X$ is *completely regular* when the following holds.\nFor every [[closed set|0ZCK8DQYN1]] $F \\subseteq X$ and every point $x \\in X \\setminus F$, there is a [[continuous function|0X3WGX95FE]] $f \\colon X \\to [0, 1]$ with $f(x) = 0$ and $f(y) = 1$ for all $y \\in F$.\n$X$ is a *Tychonoff space* when it is completely regular and [[T1|0NW6CH38YR]].\n\n## Remarks\n\nIn the definition, $T_1$ can be replaced by $T_0$, because a completely regular $T_0$ space is $T_1$.\nA Tychonoff space is [[regular|03HMJASQBQ]] and Hausdorff: the sets $f^{-1}([0, 1/2))$ and $f^{-1}((1/2, 1])$ separate $x$ from $F$.\nA space is Tychonoff if and only if it is homeomorphic to a subspace of a product $[0, 1]^I$.\nBy [[Urysohn's lemma|01EY3VR7ZD]], every [[normal|0NSR9V8TED]] $T_1$ space is Tychonoff.\n\nExample: every metric space $(X, d)$ is Tychonoff.\nFor $x \\notin F$ with $F$ closed, $d(x, F) \u003e 0$, and $f(y) = \\min\\bigl(1, d(y, x) / d(x, F)\\bigr)$ is continuous, equal to $0$ at $x$ and to $1$ on $F$.\n\nNon-example: an infinite set $X$ with the [[cofinite topology|0B906ZBTMZ]] is $T_1$ but not completely regular.\nEach continuous $f \\colon X \\to \\R$ is constant: two distinct values would have disjoint open preimages, but two nonempty open sets of $X$ always meet.\nSo no $f$ separates a point $x$ from the closed set $\\set{y}$ with $y \\ne x$.\n\nNotation: $T_{3\\frac12}$.\n",
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