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  "body": "\n## Definition\n\nLet $k$ be a real number.\nLet $M_k$ and $D_k$ be the model surface of constant curvature $k$ and its diameter bound, as for [[CAT(k) spaces|1FAF2VNA4G]].\nLet $(X, d)$ be a [[metric space|0VKD1TP7F3]].\nThe space $X$ is a *length space* if each distance $d(x, y)$ is the infimum of the lengths of continuous curves from $x$ to $y$.\nGeodesic triangles and comparison triangles in $M_k$ are as for CAT($k$) spaces.\n\nThe space $X$ is an *Alexandrov space of curvature $\\ge k$* if:\n\n1. $X$ is a [[complete|0BY52JW6NK]], [[locally compact|0FE2SB8WVJ]] length space, and\n2. each point of $X$ has a neighbourhood $U$ with the following property.\n   Take any geodesic triangle in $U$ of perimeter less than $2 D_k$, any vertex $x$ and any point $p$ on the opposite side.\n   Let $\\bar x, \\bar p$ be the comparison points in a comparison triangle in $M_k$.\n   Then $d(x, p) \\ge d_{M_k}(\\bar x, \\bar p)$.\n\nMany authors also require $X$ to have finite Hausdorff dimension; that dimension is then an integer.\n\n## Remarks\n\nExample: a complete Riemannian manifold with [[sectional curvature|1X42JAA349]] $\\ge k$ is an Alexandrov space of curvature $\\ge k$, by [[Toponogov's theorem|KX0B3AWD5H]].\nIn particular $\\R^n$ has curvature $\\ge 0$; its triangles equal their comparison triangles.\n\nNon-example: the hyperbolic plane of curvature $-1$ does not have curvature $\\ge 0$.\nTake a geodesic triangle in it and the midpoint of one side.\nThe distance from the opposite vertex to this midpoint is strictly less than in the flat comparison triangle, so clause 2 fails for $k = 0$.\n\nFor an integer $m \\ge 2$, let $C_m$ be the quotient of $\\R^2$ by the group of rotations about $0$ through multiples of $2\\pi/m$.\nThis cone is an Alexandrov space of curvature $\\ge 0$, and it is not a smooth Riemannian manifold at its tip.\n",
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