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  "body": "\n## Proof\n\nWe prove the two directions separately. The second uses the supremum.\n\n1. Recall that $S$ is [[connected|0R37DFJKFW]] when $S$ with the [[subspace topology|0D5YHQGDJY]] is a [[connected space|0D939CMME7]]. A set is open in $S$ exactly when it is $S \\cap U$ for some open $U \\subseteq \\mathbb{R}$.\n2. First, let $S$ be connected. Suppose that $S$ is not an [[interval|0TK8Y3J5RW]]. Then there are $a, b \\in S$ and $c \\notin S$ with $a \u003c c \u003c b$.\n3. The sets $(-\\infty, c)$ and $(c, \\infty)$ are open in $\\mathbb{R}$. Each point $x$ of either set has the ball of radius $|x - c|$ around it inside the set, so [[the ball criterion|0A26X79RTB]] applies.\n4. Let $U' = S \\cap (-\\infty, c)$ and $V' = S \\cap (c, \\infty)$. They are open in $S$, disjoint, and nonempty, since $a \\in U'$ and $b \\in V'$.\n5. Since $c \\notin S$, we get $U' \\cup V' = S$. So $S$ is not connected, a contradiction. Thus $S$ is an interval.\n6. Now let $S$ be an interval. Suppose that $S$ is not connected. Then there are open $U, V \\subseteq \\mathbb{R}$ with $U' = S \\cap U$ and $V' = S \\cap V$ nonempty, disjoint, and with union $S$.\n7. Choose $a \\in U'$ and $b \\in V'$. They differ, as $U' \\cap V' = \\varnothing$. By swapping the names of $U$ and $V$ if needed, we may assume $a \u003c b$.\n8. As $S$ is an interval, $[a, b] \\subseteq S$. So each point of $[a, b]$ lies in exactly one of $U$ and $V$.\n9. Let $T = U \\cap [a, b]$. It contains $a$ and is bounded above by $b$. By the [[completeness axiom|0AXWET80YW]], it has a [[supremum|0BTMQ04556]] $c$, with $a \\le c \\le b$. So $c \\in S$.\n10. Case $c \\in U$. Then $c \\neq b$, since $b \\in V$. As $U$ is open, there is $\\varepsilon \u003e 0$ with $(c - \\varepsilon, c + \\varepsilon) \\subseteq U$, and we may take $\\varepsilon \\le b - c$.\n11. Then $c + \\varepsilon / 2$ lies in $[a, b]$ and in $U$, so it lies in $T$. It is larger than $c = \\sup T$, a contradiction.\n12. Case $c \\in V$. Then $c \\neq a$, since $a \\in U$. As $V$ is open, there is $\\varepsilon \u003e 0$ with $(c - \\varepsilon, c + \\varepsilon) \\subseteq V$, and we may take $\\varepsilon \\le c - a$.\n13. By the [[approximation property of the supremum|0M8TGGP9JP]], there is $t \\in T$ with $c - \\varepsilon \u003c t \\le c$. Then $t \\in U$ and $t \\in V$, and $t \\in [a, b] \\subseteq S$.\n14. So $t \\in U' \\cap V'$, which contradicts $U' \\cap V' = \\varnothing$.\n15. Since $c \\in S = U' \\cup V'$, one of the two cases holds, and both are impossible. So $S$ is connected.\n\n## Idea\n\nA gap point $c$ outside $S$ cuts $S$ into a left part and a right part, so a connected set has no gaps. Conversely, if an interval were split into two open pieces, the supremum of the points of the first piece below a point of the second would sit on the border. Openness then shows that this border point can belong to neither piece.\n",
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