I plan to use the fact that if $f$ is monotone and non constant $[a,b]$, then it is continuous on $[a,b]$ if its range $R_f=\{f(x)|x \in [a,b]\}$ is the closed interval with endpoints $f(a)$ and $f(b)$. I dont know how to go from here.
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2In my book, "$f$ is continuous on $[a, b]$" means that $f$ is continuous at each point of $[a, b]$, so this problem is trivial. Did you misstate something? – John Hughes Oct 08 '16 at 13:49
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The idea here is this:
The fact that $f$ is monotone increasing on $[a,b]$ tells you that you have one-sided limits at every point $x\in[a,b]$. Why? Consider a sequence $(x_n)$ in $[a,b]$ that increases to the point $x$. Then $(f(x_n))$ is monotone increasing, by assumption; also, $f(x_n)$ is bounded above. (I'll let you figure out why.) So, $f(x_n)\to y$ for some $y$. Now, can you show that any other sequence $(z_n)$ which increases to $x$ must also satisfy $f(z_n)\to y$?
The same argument holds for right-handed limits. So, at every point $x_0$, there are two numbers $y_1,y_2$ such that $f(x)\to y_1$ as $x\nearrow x_0$, and $f(x)\to y_2$ as $x\searrow x_0$. It also isn't hard to see that $y_1\leq y_2$. Now, using the only assumption we haven't used yet, how can you prove that $y_1<y_2$ is impossible?
Nick Peterson
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