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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?

makansij
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    What is $D$? If it is a subset of $\mathbb{R}$, then what do we know about it? Is it connected? Closed? Open? – avs May 09 '19 at 07:47
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    Example for 2): $f(x)=x/2$ for $0<x<1$, $(2-x)$ for $1 \leq x <2$. – Kavi Rama Murthy May 09 '19 at 07:48
  • Also, I know this might sound trivial , but just to confirm, the following is NOT a counter-example, right? $f(x) =x$ for $x\in [0,0.5]$ and $f(x) =x-1$ for $x\in (1,2]$. I tried to think of an example where the domain is an interval, and it is strictly increasing, and yet the function is still continuous although it might look like it is not. – makansij May 09 '19 at 07:51
  • $D$ is a closed interval, and a subset of $\mathbb{R}$ – makansij May 09 '19 at 08:00
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    Use that for every strictly monotone function $g$, defined on an interval of the reals its inverse is continuous, see https://math.stackexchange.com/questions/605147/example-for-a-strictly-increasing-function-which-has-a-discontinuous-inverse/605219#605219. Now $g=f^{-1}$ fulfils that requirement. – Michael Hoppe May 09 '19 at 08:52
  • Thank you @KaviRamaMurthy – makansij May 09 '19 at 09:14
  • On $D=[0,1]$ let $f(0)=1, f(1)=0$ and $f(x)=x$ when $0<x<1$. The co-domain (image) of $f$ is equal to its domain $[0,1$] and $f$ is discontinuous. – DanielWainfleet May 09 '19 at 14:09

2 Answers2

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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 $f(x) = x^2$.

avs
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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 $\lim_{x\rightarrow g^+} f(x)$.

Adam Latosiński
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    Well, $D$ is not necessarily an interval, but $f(D)$ instead. – Michael Hoppe May 09 '19 at 11:51
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    @Michael Hoppe. Thanks. Fixed. – Adam Latosiński May 09 '19 at 13:21
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    You only know that $f: D \rightarrow \mathbb{R}$ is strictly increasing. From here you cannot assume the existence of a strictly increasing sequence in your first sentence: let $D={0,1}$, $f(0)=0$, $f(1)=1$. – Michael Hoppe May 09 '19 at 14:20
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    In your example $D(f)$ isn't an interval. Either way, the OP have said in a comment to the question that we're supposed to assume that $D$ is a closed interval. And even without that, we only need to prove continuty in the points in which it makes sense; if there's no strictly increasing sentence approaching $g$ then there's no such thing as $\lim_{x\rightarrow g^-}$, but that's not a problem, that's just one less condition that we need to check. – Adam Latosiński May 09 '19 at 15:28