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I'm trying to prove that if $f:[a, b]\to[s, t]$ is monotone (and its image is closed interval) then $f$ is continuous.

My attempt:

I say wlog, $f$ is increasing. I know that a monotone function only has jump discontinuities, so let's say $c\in [a, b]$ is a jump discontinuity, so:

$$K=\lim\limits_{x\to c^-} f(x)<\lim\limits_{x\to c^+} f(x)=L$$

So for each $c<x_n\to c$, $f(x_n)\to L$. $x_n>c$ so $f(x_n)\ge f(c)$ so $f(c)\le L$.

And for each $c>y_n\to c$, $f(y_n)\to K$. $y_n<c$ so $f(y_n)\le f(c)$ so $f(c)\ge K$.

So $L=K$ and this is the contradiction.

And my question - where did I use that the image of $f$ is closed interval? Did I do something wrong?

thanks?

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    $\begingroup$ Your conclusion that $L=K$ is not supported by what you have done so far. $\endgroup$ – Alex S Jul 7 '15 at 22:03
  • $\begingroup$ Oh I see.... Is this the right way though? $\endgroup$ – Stabilo Jul 7 '15 at 22:05
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    $\begingroup$ As you noticed, you need to incorporate the fact that the function's image is an interval. So there must be points in $[a,b]$ which map to points between $L$ and $K$, but how? $\endgroup$ – Alex S Jul 7 '15 at 22:07
  • $\begingroup$ Using my attempt of solution you mean, or another way? $\endgroup$ – Stabilo Jul 7 '15 at 22:09
  • $\begingroup$ After you conclude that $K\leq f(c)\leq L$, show that the existence of $K<y<f(c)$ or $f(c)<y<L$ leads to a contradiction. Conclude that $K=f(c)=L$. $\endgroup$ – Alex S Jul 7 '15 at 22:12

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