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  "body": "\n## Definition\n\nLet $X$ be a [[topological space|0CQHFPGJ45]].\nAn [[open cover|049PXYHW1T]] $\\mathcal{V}$ of $X$ *refines* an open cover $\\mathcal{U}$ if each member of $\\mathcal{V}$ lies in some member of $\\mathcal{U}$.\nThe *order* of a family $\\mathcal{V}$ is the largest $k$ such that some point of $X$ lies in $k + 1$ members of $\\mathcal{V}$.\nFor an integer $n \\ge -1$, write $\\dim X \\le n$ if every finite open cover of $X$ has a finite open refinement of order at most $n$.\nThe *Lebesgue covering dimension* $\\dim X$ is the least $n$ with $\\dim X \\le n$, and $\\dim X = \\infty$ if there is no such $n$.\n\n## Remarks\n\nThe empty space has $\\dim \\emptyset = -1$, and a nonempty finite discrete space has dimension $0$.\nThe interval $[0, 1]$ has $\\dim [0, 1] = 1$.\nGiven a finite open cover, the [[Lebesgue number lemma|040CFNJEAP]] gives a number $\\delta \u003e 0$.\nRefine the cover by finitely many relatively open intervals of length less than $\\delta$ in which only neighbours overlap.\nThis refinement has order at most $1$.\nThe cover $\\set{[0, 2/3), (1/3, 1]}$ has no open refinement of order $0$.\nSuch a refinement would be a cover by disjoint open sets, and by connectedness one of them would be all of $[0, 1]$.\nBut $[0, 1]$ lies in no member of the cover.\nSo $\\dim [0, 1] \\le 0$ fails.\nSome authors use all open covers instead of finite ones; the two versions agree for compact spaces.\nFor every $n$, $\\dim \\R^n = n$.\n\n## Explanations\n\n### L1 human\n\nTake the smallest number $k$ such that every finite open cover can be made finer with each point in at most $k$ of the sets.\nThe covering dimension of the space is $k - 1$.\n",
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