continuous function on a set of infimums

Question

The function $f:[0;1]\rightarrow\mathbb{R}$ is defined by $f(x):= \inf\{|nx-1|:n\in\mathbb{N}\}$.$\:$Show that $f$ is continuous on $(0;1]$.

First of all I'm not sure if I fully understand the function. $\:$We fix one "$x_0$" and then multiply it by every "$n$" there exists. $\:$We do that for every "$x$" in $[0;1]$.$\:$ Is it correct that the infimum is always $0$?$\:$ x is most of the time a fraction. $\:$There is always one time when "$n$" and "$x$" are inverse to each other. $\:$For example $x = \frac{1}{2}$. Then for $n=2$ we have $2*\frac{1}{2}-1 = 0$. $\:$For $x=\frac{1}{3}$ and $n=3$ $\implies 3*\frac{1}{3}-1 = 0$ and so on. $\:$So the infimum always becomes $0$. Now how to I cover the spots between the rationals?$\:$ Just apply the Intermediate value theorem or what would be the best way to do it?

Picture of function

Thanks in advance

Answer 1

Is it correct that the infimum is always 0?

Not at all. Either $f(x) = -1$ or $f(x) = x - 1$ for all $x$, depending on whether you have 0 in your $\mathbb{N}$ or not.

There is always one time when "n" and "x" are inverse to each other.

Really? What if $x = \pi$? Or $x = \frac{2}{3}$, for that matter?

So the infimum always becomes 0.

Real numbers below 0 exist.

Just apply the Intermediate value theorem or what would be the best way to do it?

You haven't proven that your function is continuous yet, so no. Also, the rationals are rather irrelevant, for both of the reasons noted above.

Given what $f$ is, it's trivial to show that it's continuous: either it's constant, or it differs from the identity by a constant, and both of those are very simple to prove to be continuous on all of [0,1]. Have you written the function down incorrectly?