Why is a strictly monotonic mapping between intervals continuous?

Solution 1:

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 $[c, A) \cup (B,d]$.

Solution 2:

So, you can have a discontinuity one of two ways: either the limit of the function at the point fails to exist (an "essential" discontinuity) or the limit does exist, but doesn't equal the value of the function at that point (a "removable" discontinuity). For a monotonically increasing function, you should be able to show that there cannot be a removable discontinuity. Then, having done this, assume that there is an essential discontinuity, and consider the greatest lower bound $\lambda$ of $f$ to the right of the point of discontinuity, and the least upper bound $\Lambda$ of $f$ to the left of it, and show that there must be a strict inequality $\lambda < \Lambda$. (Notice that each of these sets will be bounded, so $\Lambda$ and $\lambda$ are well-defined, because $f$ at any point left of the discontinuity lower-bounds $f$ at all points to the right of it, etc.) This means that the interval $(\lambda, \Lambda)$ cannot be included in the range of the function, although points above and below this interval are.

Solution 3:

Since everyone seems to be using the limit definition of continuity and I was originally taught the $\varepsilon$-$\delta$ definition, I'd like to give a proof using that. It's a direct proof, but the case analysis with the border cases is a bit ugly. Maybe that can be removed in some way? Anyway:

First we prove that $f$ is a bijection: from the given fact that $f([a,b])=[c,d]$ it follows that $f$ is surjective, and from the fact that $f$ is strictly monotonic, it follows that $f$ is injective (since any two originals mapping to the same value would contradict the strict monotonicity). So $f$ is a bijection between $[a,b]$ and $[c,d]$.

Now for $f : [a,b] \rightarrow [c,d]$ to be continuous, we need to have $$ \forall \varepsilon > 0 : \exists \delta > 0 : \forall y \in (x-\delta,x+\delta) \cap [a,b] : |f(y) - f(x)| < \varepsilon $$ for all $x \in [a,b]$. So, let an $x \in [a,b]$ and an $\varepsilon > 0$ be given.

First, assume that $x \in (a,b)$. Then the intervals $(f(x) - \varepsilon, f(x)) \cap [c,d]$ and $(f(x), f(x) + \varepsilon) \cap [c,d]$ are nonempty, because since $x \in (a,b)$, there are points less and greater than $x$ in $[a,b]$, and by the strict monotonicity points less and greater than $f(x)$ in $[c,d]$. So take some $p' \in (f(x) - \varepsilon, f(x)) \cap [c,d]$ and $q' \in (f(x), f(x) + \varepsilon) \cap [c,d]$. Since $f$ is bijective, there are unique $p,q \in [a,b]$ with $f(p) = p'$ and $f(q) = q'$, and moreover we have $p < x < q$. Now let $\delta := \min\{x-p, q-x\}$. Any point in $y \in (x-\delta,x+\delta) \cap [a,b] = (x-\delta,x+\delta)$ lies in $(p,q)$, so $f(y)$ lies in $(p',q') \subset (f(x)-\varepsilon, f(x)+\varepsilon)$ by strict monotonicity. This means that $|f(y)-f(x)| < \varepsilon$.

The choice of points in this first case is illustrated in the following picture: (here $\delta = x - p$)

Second, assume that $x = a$ (the case $x = b$ is analogous). Then from the strict monotonicity, $f(x) = c$, and since $d > c$, there is some $q' \in (c, c+\varepsilon) \cap [c,d]$. By the bijectivity of $f$, there is a unique $q \in [a,b]$ with $f(q) = q'$, and we have $a < q$. Let $\delta := q-x$; then for any point $y \in (x-\delta,x+\delta)\cap[a,b] = [x,x+\delta) = [x,q)$ (because $x = a$), we have $f(y) \in [c,q') = [f(x),q') \subset [f(x),c+\varepsilon)$ by strict monotonicity. This again means that $|f(y) - f(x)| < \varepsilon$.

Since in all cases we have $|f(y) - f(x)| < \varepsilon$, we conclude that $f$ is continuous. $\qquad\Box$