Let $f:I\rightarrow R$ be a monotonically increasing function on an open interval.If the image of this interval is an interval then would $f$ be continuous? For the case when $f(I)$ is open then I can deduct continuity of $f$.But What if $f(I)$ closed?
Continuity of a monotonically increasing function
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0I thought that that was probably the case, since $f[I]$ couldn’t be a closed interval otherwise, but I wanted to make sure, even though it doesn’t actually affect the argument. – 2012-11-29
1 Answers
Suppose that $f$ is not continuous. Since $f$ is monotone increasing, every discontinuity of $f$ must be a jump discontinuity, so there is an $a\in I$ such that $\lim\limits_{x\to a^-}f(x)<\lim\limits_{x\to a^+}f(x)$. But then $f[I]$ clearly cannot be an interval, since it contains at most one point, $f(a)$, in the interval
$\left(\lim_{x\to a^-}f(x),\lim_{x\to a^+}f(x)\right)$
but contains points on both sides of it. Thus, if $f[I]$ is an interval, $f$ must be continuous.
Added after comments: Here’s a bit more detail. Let $u=\lim\limits_{x\to a^-}f(x)$ and $v=\lim\limits_{x\to a^+}f(x)$; since $f$ is monotone non-decreasing, $u\le f(a)\le v$, and therefore at least one of the open intervals $\big(u,f(a)\big)$ and $\big(f(a),v\big)$ is non-empty. If $\big(u,f(a)\big)\ne\varnothing$, let $J=\big(u,f(a)\big)$, and otherwise let $J=\big(f(a),v\big)$; in either case $J$ is a non-empty open interval, and $J\subseteq(u,v)$. If $x, then $f(x)\le u$, and if $x>a$, then $f(x)\ge v$, so if $x\ne a$, then $f(x)\notin(u,v)$. It follows that $f[I]\cap J=\varnothing$.
$I$ is an open interval, so there are $c,d\in I$ such that $c, and therefore $f(c)\le u
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0@Pilot: I mean that $f[I]\cap\left(\lim_{x\to a^-}f(x),\lim_{x\to a^+}f(x)\right)\subseteq\{f(a)\}\;.$ Thus, either $\left(\lim_{x\to a^-}f(x),f(a)\right)$ or $\left(f(a),\lim_{x\to a^+}f(x)\right)$ must be a non-empty interval disjoint from $f[I]$. But if u and v>a, then $f(u)\le\lim_{x\to a^-}f(x)$ and $\lim_{x\to a^+}f(x)\le f(v)$, so $f[I]$ contains points on both sides of this omitted open interval and therefore cannot be an interval. – 2012-11-29