Let $f: D \rightarrow \mathbb{R}$ be strictly increasing. Show that if $f(D)$ is an interval then $f$ is continuous.

Question

I'm tasked with proving the following:

Let $f: D \rightarrow \mathbb{R}$ be strictly increasing. Show that if $f(D)$ is an interval then $f$ is continuous.

To gain some intuition for how to solve this, I tried to find an example of each of the following:

1. An example of a function which is strictly increasing, but the codomain is not an interval, and thus resulting in it not being continuous.: Let $D = [0,1]$ and let $f(x) = x $ for $x \in [0,0.5)$ and let $f(x) = 2+x$ for $x \in [0.5, 1]$.

2. An example of a function in which the codomain is an interval, but not strictly increasing, and thus resulting in it not being continuous. Having trouble coming up with an example for this one.

I saw this question: $f$ is monotone on D and $f(D)$ is an interval But to be honest it did not help me understand anything more deeply.

Can someone help me come up with an example for $2.$ ? And also give me a hint on proving this?

Answer 1

For 2, let $D = [-1, 1]$, and let $$ f(x) = \left\{ \begin{array}{lll} x + 1 & \mbox{ on } & [-1, 0[ \\ x & \mbox{ on } & [0, 1] \\ \end{array} \right. $$

Keep in mind, though, that failure of monotonicity does not necessarily result in failure of continuity; e.g., on the same $D$, take $f(x) = x^2$.

Answer 2

Hint:

Let $(x_n)_{n\in\mathbb N}$, $x_n\in D$ be a strictly increasing sequence such that $\lim_{n\rightarrow\infty} x_n = g \in D$. Since $f$ is strictly increasing, that means that $(f(x_n))_{n\in\mathbb N}$ is a strictly increasing sequence bounded from above by $f(g)$. Therefore $(f(x_n))_{n\in\mathbb N}$ is convergent and $\lim_{n\rightarrow\infty} f(x_n) \le f(g)$. Assume that $\lim_{n\rightarrow\infty} f(x_n) < f(g)$ and show a contradiction with $f(D)$ being an interval. Indirectly you'll prove that $\lim_{x\rightarrow g^-} f(x) = g$. Then do an analogous thing for $\lim_{x\rightarrow g^+} f(x)$.