Is the following proof of the continuity of a function correct?

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.

• 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 $|x_0-x|+\inf\{|x-k|\} < \inf\{|x_0 -k|\} \le \inf\{|x_0-x| + |x-k|\}$? Commented Feb 16, 2021 at 0:38

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 $$x, y\in \mathbb{R}$$.