# Strictly increasing continuous function

Prove that any onto strictly increasing map $f: (0,1) \to (0,1)$ is continuous.

Since its strictly increasing then for $x<y$ it implies that $f(x) < f(y)$. For continuity I must show that for any $y\in (0,1)$ there exists a $\delta>0$ such that for $\epsilon>0$ then $|x - y|<\delta$ implies that $|f(x) - f(y)|<\epsilon.$

• "Prove that any onto strictly increasing map is continuous." could be constued as "Pick any onto strictly increasing map and prove that it is continuous." But I don't think that's what is meant. Changing "any" to "every" removes all ambiguity. In this particular instance, misunderstandings may not follow, but if you're writing about something complicated that your readers have to work hard to understand, they might. – Michael Hardy Nov 16 '13 at 2:57
• There's a really simple way to find $\delta$ when given $\varepsilon$. My answer below explains how. – Michael Hardy Nov 16 '13 at 3:01

Let $a\in(0,1)$.

Let $\varepsilon>0$.

Since $f$ is onto, for some point $b$ we have $f(b)=f(a)-\varepsilon$, and for some point $c$ we have $f(c)=f(a)+\varepsilon$.

Let $\delta=\min\{c-a,b-c\}$.

Then prove that that value of $\delta$ is small enough.

• Can you elaborate? I want to know why you chose $f(b)=f(a)-\varepsilon$ and $f(c)=f(a)+\varepsilon$. – user104235 Nov 17 '13 at 2:55
• I chose those precisely because the whole idea of an $\varepsilon$-$\delta$ proof of continuity is to make $f(x)$ lie between those two numbers. That can be done by making $x$ lie between $b$ and $c$. So we want to say: if $x$ lies between $a-\delta$ and $a+\delta$, then $x$ lies between $b$ and $c$. Then we just ask which of $b$ and $c$ is closer to $a$, and that distance is $\delta$. If the distance between $a$ and $x$ is less than $\delta$, then $x$ is between $b$ and $c$, so $f(x)$ is between $f(a)-\varepsilon$ and $f(a)+\varepsilon$. – Michael Hardy Nov 17 '13 at 2:59
• Michael, I am trying to do it your way can you help me finish? – user104235 Nov 18 '13 at 2:18
• I tried to do find a contraction so I let $p \in (0,1)$ then for $\epsilon>0$ suppose no delta exists $\delta>0$ such that $|x-p|<\delta$ implies $|f(x) - f(p)|<\epsilon$. Let $\delta = 1/n$ and $x_n\in(p-1/n,p+1/n)$ such that $|f(x_n) - f(p)|> \epsilon$. Suppose $x_n <p$ then $f(x_n) < f(p)$ hence $f(p) - f(x_n)>\epsilon$. – user104235 Nov 18 '13 at 2:22
• Depends on what you're trying to do. We've got $b<a<c$. The distance from $a$ to $b$ is $a-b$. The distance from $a$ to $c$ is $c-a$. Call the smaller of those two $\delta$. Then $b\le a-\delta<a+\delta\le c$. Consequently, if $x$ is between $a-\delta$ and $a+\delta$, then $x$ is between $b$ and $c$, so $f(x)$ is between $f(a)-\varepsilon$ and $f(a)+\varepsilon$. So that proves that given $\varepsilon>0$, there exists a suitable $\delta>0$. ${}\qquad{}$ – Michael Hardy Nov 18 '13 at 2:32

Hint: In this case, it's quite a bit easier to use the (equivalent) definition of continuity that

$f$ is continuous if and only if for all open sets $O$, $f^{-1}(O)$ is open.

Furthermore, it's sufficient to show that $f^{-1}((\alpha, \beta))$ is open for every interval $(\alpha, \beta)$; use the fact that $f$ is strictly increasing to show that

$$f^{-1}((\alpha, \beta)) = (f^{-1}(\alpha), f^{-1}(\beta))$$

• This is easier after you know the definition of open sets and after you've proved that the $\varepsilon$-$\delta$ definition is equivalent to this one. Doing those things is worthwhile because it makes easier proofs like this possible. But there's a really simple way to do it that can be understood if one knows ONLY the $\varepsilon$-$\delta$ definition. See my posted answer here. – Michael Hardy Nov 16 '13 at 3:03
• I read a theorem just now that shows me how to do it this way, thank you, . – user104235 Nov 17 '13 at 3:05