$I$ and $J$ are intervals.

My attempt:

Let $x_0 \in I$ and $\varepsilon > 0$.

Suppose $x_0$ is an interior point of $I$. Then $f(x_0)$ is an interior point of $J$ and there exists $x_1,x_2\in I$ such that:

$$f(x_0)-\varepsilon < f(x_1) < f(x_0) < f(x_2) < f(x_0)+\varepsilon $$

Since $f$ is monotone, we have $x_1 = x_0 - \delta_1, x_2 = x_0 + \delta_2, \delta_1, \delta_2 >0$

Take $\delta = \min \{\delta_1,\delta_2\}$ and we have:

$$x\in (x_0 - \delta,x_0 + \delta) \cap I = (x_0 - \delta,x_0 + \delta) \Rightarrow f(x_1)<f(x)<f(x_2)$$ $$ \Rightarrow f(x)\in(f(x_0)-\varepsilon,f(x_0)+\varepsilon)$$

Thus $f$ is continuous.

Now suppose that $x_0$ is not an interior point of $I$.

Then there exists $x_1 \in I$ such that $f(x_0)-\varepsilon < f(x_1) < f(x_0) $ or $f(x_0) < f(x_1) < f(x_0)+\varepsilon $.

Suppose that $f(x_0)-\varepsilon < f(x_1) < f(x_0) $. Then we have $x_1 = x_0 - \delta$. Thus:

$$x\in (x_0 - \delta,x_0 + \delta) \cap I = (x_0 - \delta,x_0] \Rightarrow f(x_1)<f(x)\leq f(x_0)$$ $$ \Rightarrow f(x)\in(f(x_0)-\varepsilon,f(x_0)+\varepsilon)$$

Thus $f$ is continuous.

If $f(x_0) < f(x_1) < f(x_0)+\varepsilon $ we can proceed in a similar way.

Does it look right? Is there an easier way to solve this problem?

  • $\begingroup$ This is a bit tedious but the proof is correct and rigorous. Note that you only used the fact that $f$ is monotone and surjective (and $I$ and $J$ are intervals). I think the main thing that can be improved is the explicit mention of these properties where they are useful (and necessary). $\endgroup$ – nombre Dec 26 '15 at 22:43
  • $\begingroup$ If I feel like being pedantic I could point out that the empty set and a singleton set $\{x\}$, with $x\in R$ are intervals also, but for your Q they are trivial cases. Good proof. $\endgroup$ – DanielWainfleet Dec 26 '15 at 23:22

If $f$ is monotone increasing and a bijection then for any interval $(a,b) \subseteq J$, as $f^{-1}$ is also a monotone-increasing bijection, $f^{-1}((a,b))$ must be some interval $(c,d) \subseteq I$. Thus the inverse image of any open set is open, so $f$ is continuous.


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