[Math] Why is a strictly monotonic mapping between intervals continuous

continuitymonotone-functionsreal-analysis

Consider the following problem:

Say $a,b\in\mathbb R$, $a <b$ and $f:[a,b]\rightarrow\mathbb R $ strictly monotonic, let's say increasing for the sake of simplicity. Let $f ([a,b])=[c,d]$ for some $c,d\in\mathbb R$. Show that $f$ is continuous.

This question is part of an old exam in introductory real analysis. Intuitively, I understand that a discontinuity in $f$ would imply a "breach" in the interval of the image. I'm having trouble with a rigorous proof, however.

Edit:
One of the comments on this question suggests a proof that involves upper and lower limits. As I am not very comfortable with those, I was wondering if there is a (simple) proof that abstains from their usage.

Best Answer

Direct proof. Suppose the function $f:[a,b]\rightarrow \mathbb{R}$ is increasing, with $f([a,b])=[c,d]$. Pick $x$ and let $S$ be any sequence which converges to $x$. It suffices to show that the image of $S$ converges to $f(x)$.

Let $N_1 \supseteq N_2 \supseteq N_3 \supseteq \cdots $ be a strictly nested sequence of open intervals around $f(x)$ and moreover suppose the length of the intervals converges to 0.

  1. Note that the inverse images are nested, too: $f^{-1}(N_i) \supseteq f^{-1}(N_{i+1})$. (Think about the how the endpoints of the nested intervals $N_i$ are $<$ related to each other.)

  2. Here's the clever part. Because $f([a,b]) = [c,d]$, it must follow that each inverse image $f^{-1}(N_i)$ contains an open interval $M_i$ around $x$.

    Indeed, for contradiction, let $N_i = (p,q)$ be an interval around $f(x)$ whose inverse image contains no open interval. Then the monotonicity of $f$ therefore implies that $$\begin{cases}f(y) < p & \forall y < x\\ q < f(y) & \forall y > x \end{cases}.$$ This contradicts our assumption that $f([a,b])$ is a complete interval, since $f^{-1}([p,q]) = \{x\}$. In other words, we would arrive at a contradiction that $f([a,b])$ has at least one gap in it, something like $[c\cdots p)\cdots f(x)\cdots (q\cdots d]$.

  3. Because the sequence $S$ converges to $x$, and the $M_i$ contain open intervals around $x$, for each such open interval $M_i$, the tail of the sequence $S$ eventually lies entirely within $M_i$.

  4. Hence the image of the sequence $S$ eventually lies entirely within each $N_i$.

  5. But we assumed that the lengths of the $|N_i|$ converge to zero; hence the image of the sequence $S$ converges to $f(x)$.

Proof by contradiction. Let's assume that the increasing function $f:[a,b]\rightarrow \mathbb{R}$, $f([a,b])=[c,d]$, is discontinuous and arrive at a contradiction.

If $f$ is discontinuous at some point $p\in[a,b]$, this means that either the limit and function value are different at $p$, or the limit doesn't exist at all. But the limit and function value can't be different, because that contradicts the fact that $f$ is monotone.

Why? Assume without loss of generality that the function value is greater than the limit value. Measure the distance between the function value $q\equiv f(p)$ and the limit value $q^\prime = \lim_{x\rightarrow p}f(x)$; for safety, let $\epsilon$ be a quarter of that distance. By definition of limit, there is a neighborhood around $p$ whose points all map within $\epsilon$ distance of the limit $q^\prime$. In particular, there are points $p_1 < p < p_2$ that do so— $f(p_1)$ and $f(p_2)$ are within $\epsilon$ distance of $q^\prime$. This means that $q^\prime - \epsilon < f(p_1)$, $f(p_2) < q^\prime + \epsilon$. Because $f$ is increasing, then we should more generally have

$$q^\prime - \epsilon < f(x) < q^\prime + \epsilon$$

for every $x\in [p_1, p_2]$. But $p$ itself violates this rule. Remember that $f(p) = q^\prime + 4\epsilon$. Hence we find

$$q^\prime - \epsilon < f(p) = q^\prime + 4\epsilon \not< q^\prime + \epsilon.$$

This is a contradiction because $p < p_2$ but $f(p) > f(p_2)$. (If we had assumed the other case where the function value was less than the limit value, we would instead have gotten a contradiction from $p_1 < p$ but $f(p_1) > f(p)$.)

So the limit doesn't exist at all. Let $(a_n)$ be a strictly increasing sequence $a_1 < a_2 < a_3 < \cdots$ whose limit is $p$. Let $b_1 > b_2 > b_3 > \ldots$ be a strictly decreasing sequence whose limit is also $p$. (Thus $a_i$ approaches $p$ from below, and $b_i$ approaches $p$ from above.)

By monotonicity, $f(a_1) \leq f(a_2) \leq f(a_3) \leq \cdots$, and also $f(b_1) \geq f(b_2) \geq f(b_3) \geq \cdots$. We can consider the limit $A$ and $B$ of these sequences.

First, note that the limits $A$ and $B$ must exist because the sequences are monotonic and their values are all bounded between $f(a_1)$ and $f(b_1)$. Next note that since $f(a_i) \leq f(b_j)$ for all $i$ and $j$, the limit values must have $A \leq B$. Finally, note that we can exclude the case $A = B$, because if $A=B$ then $f$ is continuous at $p$, violating our definition of $p$. Hence $A$ is strictly less than $B$.

By monotonicity again,

$$\begin{cases} f(x) \leq A & x < p\\ f(x) \geq B & x > p. \end{cases}$$

This implies that there are no values $f(x)$ in the interval $[A,B]$. That's a contradiction, since we expected the image of the domain $[a,b]$ to be the complete interval $[c,d]$, not just some subset $[c, A) \cup (B,d]$.

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