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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

the_asdf_word
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1 Answers1

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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?

user3482749
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  • thanks for your reply. I added a picture of the function. I couldn't see a typo from my side – the_asdf_word Nov 15 '18 at 17:30
  • Ah, that makes a little more sense. But still, for which natural number $n$ is $\frac{2}{3}n = 1$? – user3482749 Nov 15 '18 at 17:32
  • yeah ok that's not possible. What would be the greater idea how to prove it? – the_asdf_word Nov 15 '18 at 20:08
  • So, for all $x$ in $(0,\frac{1}{2}]$, we have $f(x) = x$ (you should prove this). This is clearly continuous. For $x$ in $(\frac{1}{2},1]$, we have $f(x) = 1 - x$ (prove this too). This is clearly continuous. Then all you need to do is check that the two match up at $\frac{1}{2}$, which they do, since $\frac{1}{2} = 1 - \frac{1}{2}$. – user3482749 Nov 15 '18 at 21:46
  • I'm sorry I don't see it. say x = 0.3. Then $x\in(0;\frac{1}{2}]$ but $|3*0.3-1|=0.1$ which is a way better infimum and it's not x – the_asdf_word Nov 16 '18 at 00:31