{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe check the second form of the definition of a [[basis|06EX0FHBTF]] for both families.\n\n1. Let $\\mathcal{B}$ be the set of all [[open balls|0DC3MPR7C9]] $B(x, r)$ with $x \\in X$ and $r \u003e 0$. Let $\\mathcal{B}'$ be the set of balls $B(x, 1/n)$ with $x \\in X$ and integer $n \\ge 1$. Then $\\mathcal{B}' \\subseteq \\mathcal{B}$.\n2. By [[open balls are open|0B2305GNAZ]], each member of $\\mathcal{B}$ is in the [[metric topology|0A6YRNZB2R]] $\\tau_d$. So $\\mathcal{B}' \\subseteq \\mathcal{B} \\subseteq \\tau_d$.\n3. Let $U \\in \\tau_d$ and $x \\in U$. By the definition of $\\tau_d$, there is a real $\\varepsilon \u003e 0$ with $B(x, \\varepsilon) \\subseteq U$.\n4. The ball $B(x, \\varepsilon)$ is in $\\mathcal{B}$ and contains $x$, since $d(x, x) = 0 \u003c \\varepsilon$. So $\\mathcal{B}$ is a basis of $\\tau_d$.\n5. Since [[the reals are Archimedean|0A3GWZ86AY]], there is an integer $n \\ge 1$ with $n \\varepsilon \u003e 1$, that is, $1/n \u003c \\varepsilon$.\n6. If $d(x, y) \u003c 1/n$, then $d(x, y) \u003c \\varepsilon$. So $x \\in B(x, 1/n) \\subseteq B(x, \\varepsilon) \\subseteq U$.\n7. The ball $B(x, 1/n)$ is in $\\mathcal{B}'$. So $\\mathcal{B}'$ is also a basis of $\\tau_d$.\n\n## Idea\n\nBalls are open, and by the very definition of the metric topology every open set contains a ball around each of its points. A ball of radius $1/n$ fits inside any given ball with the same centre once $n$ is large.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "54E35"
    ],
    "name": "Proof that open balls form a basis of the metric topology",
    "proves": "02EEX96DFS",
    "sources": [
      {
        "locator": "Examples",
        "url": "https://en.wikipedia.org/wiki/Base_(topology)"
      }
    ],
    "tag": "TKNNQY6RF9",
    "uses": [
      "06EX0FHBTF",
      "0A6YRNZB2R",
      "0B2305GNAZ",
      "0DC3MPR7C9",
      "0A3GWZ86AY"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/TKNNQY6RF9/",
  "links": [
    {
      "field": "proves",
      "target": "02EEX96DFS"
    },
    {
      "field": "uses",
      "target": "06EX0FHBTF"
    },
    {
      "field": "uses",
      "target": "0A6YRNZB2R"
    },
    {
      "field": "uses",
      "target": "0B2305GNAZ"
    },
    {
      "field": "uses",
      "target": "0DC3MPR7C9"
    },
    {
      "field": "uses",
      "target": "0A3GWZ86AY"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/TKNNQY6RF9.md",
  "msc": [
    "54E35"
  ],
  "name": "Proof that open balls form a basis of the metric topology",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/TKN/TKNNQY6RF9.md",
  "status": null,
  "tag": "TKNNQY6RF9",
  "uri": "m2://TKNNQY6RF9",
  "url": "https://math2.tamnd.com/wiki/proof-that-open-balls-form-a-basis-of-the-metric-topology/",
  "writing": {
    "tag": "TKNNQY6RF9",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}