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  "body": "\n## Statement\n\nLet $(X, \\tau)$ be a [[topological space|0CNQXVBHCX]] and let $x \\in X$.\nA set $N \\subseteq X$ is a neighbourhood of $x$ when there is an [[open set|01ZQQDCERP]] $U$ with $x \\in U \\subseteq N$.\nThen a set $A \\subseteq X$ is open if and only if $A$ is a neighbourhood of each of its points.\n\n## Remarks\n\nProof of the converse: for each $x \\in A$, choose an open $U_x$ with $x \\in U_x \\subseteq A$. Then $A = \\bigcup_{x \\in A} U_x$, which is open as a [[union|05GK32H72H]] of open sets.\nA neighbourhood need not be open. For example, $[-1, 1]$ is a neighbourhood of $0$ in $\\mathbb{R}$.\nIn a metric space, $N$ is a neighbourhood of $x$ if and only if $N$ contains a ball $B(x, \\varepsilon)$ for some $\\varepsilon \u003e 0$.\n\n## Explanations\n\n### L1 human\n\nA neighbourhood of $x$ is a set that contains all points near enough to $x$.\nA set is open when it is a neighbourhood of each of its points. Then no point of the set lies on its edge.\n",
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        "locator": "Neighbourhood of a point",
        "url": "https://en.wikipedia.org/wiki/Neighbourhood_(mathematics)"
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        "locator": "Topology from neighbourhoods",
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  "msc": [
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  "name": "Open sets characterised by neighbourhoods",
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  "url": "https://math2.tamnd.com/wiki/open-sets-characterised-by-neighbourhoods/",
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