6
$\begingroup$

I just started learning analysis and encountered my first task on continuity today, namely proving that $f: \mathbb{R}\to \mathbb{R}, \space f(x) = \inf\{|x-k\rvert \space \mid k\in \mathbb{Z} \}$ is continuous in every point. I attempted to prove this as follows:

Let $x_{0} \in \mathbb{R}$ be arbitrary. Let $x \in \mathbb{R}$. By the triangle inequality we have $\left|x_{0}-k\right| \leq\left|x_{0}-x\right|+|x-k|$ for all $k \in \mathbb{Z}$. Thus, one obtains $$ \inf \left\{\left|x_{0}-k\right| \mid k \in \mathbb{Z}\right\} \leq\left|x_{0}-x\right|+\inf \{|x-k| \mid k \in \mathbb{Z}\} $$ so by the definition of $f$ $$ f\left(x_{0}\right)-f(x) \leq\left|x_{0}-x\right| $$ so we obtain $f(x)-f\left(x_{0}\right) \leq$ $\left|x_{0}-x\right|$ (applying the same argumentation). We get $$ \left|f\left(x_{0}\right)-f(x)\right| \leq\left|x_{0}-x\right| $$ Now, let $\epsilon>0$ be arbitrary. To show that $f$ is continuous in $x_{0}$ we must find some $\delta>0$, s.t. for all $x \in \mathbb{R}$ with $\left|x-x_{0}\right|<\delta$ the inequality $\left|f\left(x_{0}\right)-f(x)\right|<\epsilon$ holds. From the above we can clearly see that $\delta=\epsilon$ is a sufficient choice is. Since $x_{0}$ was arbitrary, we are done.

Does this attempt seem valid? Also, what are other ways one could prove this (assuming that the above works). I assume that one often can just plug in the definition of the respective function and then rearrange the inequality, but this didn‘t quite work for me in this case.

$\endgroup$
1
  • $\begingroup$ But note $\inf\{|x_0-x|\}+ \inf\{|x-k|\}\le \inf\{|x_0-x|+|x-k|\}$ .(It's possible that the $x$ and $k$ that make$\inf\{|x_0-x|+|x-k|\}$the minimum might be a different $x$ then makes $|x_0-x|$ minimal and a different $x,k$ that makes $\inf\{|x-k|\}$ minimal.) What i fthere is an $x$where $|x_0-x| +\inf\{|x-k|\}< \inf\{|x_0-x| + |x-k|\}$ and why can′t we have $|x_0-x|+\inf\{|x-k|\} < \inf\{|x_0 -k|\} \le \inf\{|x_0-x| + |x-k|\}$? $\endgroup$
    – fleablood
    Feb 16 at 0:38
2
$\begingroup$

Your estimate appears to be correct. If you draw a picture, this function is a "sawtooth" function. For $n\in \mathbb{Z}$, you have $f(n) = 0$ and $f(n + 1/2) = 1/2$. The rest of the graph can be obtained by "connecting the dots." The slopes of the connecting lines are $\pm 1$. So, you do have $|f(x) - f(y)| \le |x - y|$ for all $x, y\in \mathbb{R}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.