# Prob. 16, Chap. 4 in Baby Rudin: Discontinuities of the bracket function

Here is Prob. 16, Chap. 4 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Let $[x]$ denote the largest integer contained in $x$, that is, $[x]$ is the integer such that $x-1 < [x] \leq x$; and let $(x) = x- [x]$ denote the fractional part of $x$. What discontinuities do the functions $[x]$ and $(x)$ have?

My effort:

Let $f$ and $g$ be the mappings of $\mathbb{R}^1$ into $\mathbb{R}^1$ defined by $$f(x) = [x] \ \mbox{ for all } x \in \mathbb{R}^1,$$ and $$g(x) = (x) \ \mbox{ for all } x \in \mathbb{R}^1.$$ Now suppose that $p$ is a real number that is not an integer, and let $\varepsilon > 0$ be a given real number, let us take $\delta$ such that $$0 < \delta < \frac{1}{2} \min \left( p-[p], [p]+1-p \right).$$ Then, for any real number $x$ such that $\vert x-p \vert < \delta$, we have $$[p] < x < [p]+1,$$ and note that $$\vert f(x) - f(p) \vert = \vert [p]-[p]\vert = 0 < \varepsilon,$$ showing that $f$ is continuous at any real number $p$ that is not an integer.

Now suppose that $p$ is an integer. Then if $0< \delta < 1$, we note that, for any real number $x$ such that $p \leq x < p+\delta$, the following is true. $$\vert f(x) - f(p) \vert = \vert [p] - [p] \vert = 0 < \varepsilon,$$ which shows that $$\lim_{x \to p+0} f(x) = f(p),$$ that is, $f$ is continuous from the right at $p$.

On the other hand, if $p-\delta < x < p$, then we note that then $p-1 < x < p$ and so $$\vert f(x) - f(p) \vert = \vert [p]-1 - [p] \vert = 1 \not< \varepsilon$$ if $0 < \varepsilon \leq 1$, thus showing that $f$ is not continuous (from the left) at $p$.

Now let $i$ be the identity mapping of $\mathbb{R}^1$ into $\mathbb{R}^1$. Then $i$ is (uniformly) continuous on (all of) $\mathbb{R}^1$. Moreover $g = i-f$ and so $g$ is continuous on the set $\mathbb{R} - \mathbb{Z}$ of real numbers other than integers and is continuous from the right at every integer $p$.

However, if $p$ is an integer and if we take any $\delta$ such that $0 < \delta < \frac{1}{2}$, then we see that if $p-\delta < x < p$, then $p-1 < x < p$ and so $$\vert g(x) - g(p) \vert = \vert \left( x-[x]\right) - \left( p - [p] \right) \vert = \vert \left( x-[p]-1 \right) - \left( p - [p] \right) \vert = \vert x-1 - p \vert \geq 1 - \vert x-p \vert > \frac{1}{2} \not< \varepsilon$$ for any real number $\varepsilon$ such that $0 < \varepsilon \leq \frac{1}{2}$, thus showing that $g$ is not continuous (from the left) at any integer $p$.

Is the above solution correct? If so, are my proofs rigorous enough?

Or, is there any flaw in my reasoning?

• The part that begins with "suppose $p$ is a real number that is not an integer" ends with "showing that $f$ is continuous at any integer $p$", which is ultimately false. – user228113 Feb 2 '17 at 10:19
• As @G.Sassatelli pointed out, you are missing a non-integral at that very point. Besides that, your proofs are correct. – b00n heT Feb 2 '17 at 10:20