2
$\begingroup$

Let $f: [a,b]\rightarrow [c,d]$ be a surjective and strictly increasing function. Show that $f$ is continuous.


I have already proved that any function $f$ in $\mathbb{R}$ is continuous if and only if the preimage of a closed set is again a closed set. Thus it is enough to prove that if $x\le d$ then $f^{-1}([c,y])$ is closed or that $f^{-1}([c,y])=[a,f^{-1}(\{y\})]$ (using the fact that $f$ is surjective and strictly increasing - thus injective so that $f^{-1}(\{y\})$ is exactly one element from $[a,b]$).

Choose $y\in (c,d]$. Assume that there is $p \in [a,f^{-1}(\{y\})]$ and $f(p)>y$. Since $f$ is strictly increasing we have $p>f^{-1}(\{y\})$. Contradiction.

Could someone "verify" this proof? I don't think this is correct. Thanks.

$\endgroup$
  • $\begingroup$ But why is it enough to prove that the preimage of any closed subinterval is closed? Are intervals the only examples of closed subsets of $[c,d]$? $\endgroup$ – layman Dec 6 '15 at 12:31
  • 1
    $\begingroup$ Yes, that is my problem. I forgot about this;/ $\endgroup$ – luka5z Dec 6 '15 at 12:36
  • $\begingroup$ $x_m < x_0 < x_n \Rightarrow f(x_m) <f(x_0) <f(x_n)$; take sequences { $x_m$ } and { x_n$ } respectively increasing and decreasing, both convergent to $x_0$. There can be no jumps because f is surjective. $\endgroup$ – Piquito Dec 6 '15 at 13:46
  • $\begingroup$ @Ataulfo I don't quite understand what do you mean. Could you post an answer and explain? $\endgroup$ – luka5z Dec 6 '15 at 14:08
  • $\begingroup$ I mean for all convergent sequence to x the corresponding images by f convergent to f(x) which is a definition of continuity in a metric space $\endgroup$ – Piquito Dec 7 '15 at 11:40
2
$\begingroup$

Step 1. Suppose $f$ is not continuous at some point $x \in [a,b]$. Then this means there is some $\epsilon > 0$ so that for all $\delta > 0$ we can find $y$ with $|x - y|<\delta$ and $|f(x) - f(y)| \geq \epsilon$.

Step 2. Ok, well, that means for $\delta = \frac{1}{n}$, we can find $y_{n}$ with $|x - y_{n}| < \frac{1}{n}$ and $|f(x) - f(y_{n})| \geq \epsilon$. In particular, we can assume for all $n$, $x < y_{n+1} < y_{n}$ by choosing $y_{n}$ from the interval $(x, \delta_{n}')$ in each step, where $\delta_{n}' = \min\{ y_{n -1}, \frac{1}{n}\}$.

Step 3. Ok, so we have $f(y_{n+1}) < f(y_{n})$ since $f$ is strictly increasing. But for all $n$, we have $|f(x) - f(y_{n})| \geq \epsilon$, i.e., $f(y_{n}) - f(x) \geq \epsilon$ (since $f$ is increasing and $x < y_{n}$). Thus, $f(y_{n}) \geq f(x) + \epsilon$ for all $n$, with $y_{n}$ decreasing down to $x$.

Step 4. Now, since $f$ must be onto, there must be some elements mapped into $(f(x), f(x) + \epsilon)$. But we are about to show that no elements can be mapped into this interval. Here is how:

Case 1: $y > x$. Since $y_{n} \to x$, that means if $y > x$, then $y > y_{n}$ for some $n$, so $y \not \in (f(x), f(x) + \epsilon)$.

Case 2: $y < x$. If $y < x$, then since $f$ is strictly increasing, $f(y) < f(x)$, so we can't have $f(y) \in (f(x), f(x) + \epsilon)$.

Thus, in every possible case, no elements are mapped into $(f(x), f(x) + \epsilon)$, which contradicts that $f$ is onto.

$\endgroup$
1
$\begingroup$

I try to resist zombie threads but:

I find it useful to rely on common sense understanding first. And then formalization.

So common sense: surjective means every point is mapped to. Strictly increasing means every value is larger than the previous values. Continuous means it does not jump.

Since every point is mapped to, and it must pass through every value and omit any. And as it is increasing any jump would be forward and it can not go back to the points it skipped over. So there can be no jumps. So it must be continuous.

Okay, formalize it: Surjective and strictly increasing implies continuous.

Continuous (def): At every point $e \in [a,b]$ for every $\epsilon > 0$ there is a $\delta $ so that $|x-e|< \delta \implies |f(x) - f(x)| < \epsilon$.

Surjective: for any $y\in [c,d]$ there is an $x \in [a,b]$ so that $f(x) = y$.

So for any $a < e < b$ we have $c \le f(a) < f(e) < f(b)\le d$ and for any $\epsilon > 0$ let $ k = \max(f(e) - \epsilon, f(a))$ and $j = \min(f(e) + \epsilon, f(b))$. As $f$ is surjective there exist $k', j'$ so that $f(k') =k; f(j') = j$ and as $f$ is strictly increasing $k' < e < j'$.

Let $\delta = \min(e - k', j' - e)$

If $|x - e| < \delta$ then $k' < x < j'$ and $f(e) - \epsilon \le f(k') = k < f(x) < f(j') = j \le f(e) + \epsilon $. So $|f(e) - f(x) | < \epsilon$.

So $f$ is continuous.

Which was.... surprisingly tedious, despite being intuitively obvious.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.