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  "body": "\n## Proof\n\nHere $f$ is increasing in the sense of a [[monotone function|0WFF6AZPBP]]: $x \\le y$ implies $f(x) \\le f(y)$.\n\n1. Let $S = \\set{f(x) : a \u003c x \u003c c}$. It is nonempty, since $\\frac{a + c}{2} \\in (a, c)$.\n2. For $a \u003c x \u003c c$ we have $f(x) \\le f(c)$, since $f$ is increasing. So $f(c)$ is an upper bound of $S$.\n3. By the [[completeness axiom|0AXWET80YW]], $S$ has a [[supremum|0BTMQ04556]] $L$. Since $f(c)$ is an upper bound, $L \\le f(c)$.\n4. The point $c$ is a [[limit point|07EGCNHAVW]] of $(a, b) \\cap (-\\infty, c) = (a, c)$. Indeed, for $\\delta \u003e 0$ the point $c - \\frac{1}{2}\\min(\\delta, c - a)$ lies in $(a, c)$ and within $\\delta$ of $c$.\n5. Let $\\varepsilon \u003e 0$. The number $L - \\varepsilon$ is not an upper bound of $S$. So there is $x_0 \\in (a, c)$ with $f(x_0) \u003e L - \\varepsilon$.\n6. Put $\\delta = c - x_0 \u003e 0$. Let $x \\in (a, b)$ with $c - \\delta \u003c x \u003c c$. Then $x_0 \u003c x \u003c c$.\n7. Since $f$ is increasing, $L - \\varepsilon \u003c f(x_0) \\le f(x)$. Since $x \\in (a, c)$, also $f(x) \\le L$. So $\\abs{f(x) - L} \u003c \\varepsilon$.\n8. By the definition of the [[one-sided limit|0909PWZR62]], steps 4 to 7 give $\\lim_{x \\to c^-} f(x) = L = \\sup_{a \u003c x \u003c c} f(x)$.\n9. Now let $T = \\set{f(x) : c \u003c x \u003c b}$. It is nonempty, and $f(c)$ is a lower bound of $T$, since $f$ is increasing.\n10. By the [[existence of the infimum|0EDH403M21]], $T$ has an [[infimum|04KFTA34C5]] $R$. Since $f(c)$ is a lower bound, $f(c) \\le R$.\n11. In the same way as step 4, $c$ is a limit point of $(c, b)$.\n12. Let $\\varepsilon \u003e 0$. The number $R + \\varepsilon$ is not a lower bound of $T$. So there is $x_1 \\in (c, b)$ with $f(x_1) \u003c R + \\varepsilon$.\n13. Put $\\delta = x_1 - c \u003e 0$. Let $x \\in (a, b)$ with $c \u003c x \u003c c + \\delta$. Then $c \u003c x \u003c x_1$, so $R \\le f(x) \\le f(x_1) \u003c R + \\varepsilon$. Thus $\\abs{f(x) - R} \u003c \\varepsilon$.\n14. So $\\lim_{x \\to c^+} f(x) = R = \\inf_{c \u003c x \u003c b} f(x)$.\n15. Steps 3, 8, 10 and 14 give the chain $\\lim_{x \\to c^-} f(x) = L \\le f(c) \\le R = \\lim_{x \\to c^+} f(x)$.\n\n## Idea\n\nTo the left of $c$, the values of $f$ rise but never pass $f(c)$, so they have a least upper bound. Once one value comes within $\\varepsilon$ of that bound, all later values up to $c$ stay within $\\varepsilon$, because $f$ is increasing. The right side is the mirror image with the infimum.\n",
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