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4.3 Discontinuities. Variation of a theme

Proposition 4.3.1

There exists a function $f:\mathbb{R}\to\mathbb{R}$ such that $f$ is not continuous at ANY $x\in\mathbb{R}$ .

Proof: Let $f$ be defined the following (completely legitimate) way.

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Let $f(x)=0$ if $x$ is a rational number.
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Let $f(x)=1$ if $x$ is an irrational number.

If $x\in\mathbb{R}$ , then there exists a sequence of rational numbers $\{q_{n}\}^{\infty}_{n=1}$ tending to $x$ , that is $f(q_{n})\to 0$ . Also, there exists a sequence of irrational numbers $\{r_{n}\}^{\infty}_{n=1}$ tending to $x$ , that is $f(r_{n})\to 1$ . Therefore by the Sequential Definition, $f$ cannot be continuous at $x$ . $\square$

Now we make a small twist of this trick.

Proposition 4.3.2

There exists a function $f:\mathbb{R}\to\mathbb{R}$ such that $f$ is continuous at one single point.

Let $g(x)=f(x)x^{2}$ , where $f$ is the function in the previous proposition. Then we “ruined” the discontinuity at zero! Indeed, if $x_{n}\to 0$ , then $f(x_{n})x_{n}^{2}\to 0$ as well, since the sequence $\{f(x_{n})\}^{\infty}_{n=1}$ is bounded and $x_{n}^{2}\to 0$ (Proposition 2.4.1). On the other hand, we have not ruined the discontinuity at any other point. In fact, if $x\neq 0$ , then we will have a convergent sequence of irrational numbers $r_{n}\to x$ such that $g(r_{n})\to x^{2}>0$ and a convergent sequence of rational numbers $q_{n}\to x$ such that $g(q_{n})\to 0$ . $\square$

Proposition 4.3.3

There exists a function $f:\mathbb{R}\to\mathbb{R}$ such that $f$ is continuous at $x\in\mathbb{R}$ if and only if $x$ is an irrational number.

Note: There exists NO function $f:\mathbb{R}\to\mathbb{R}$ such that $f$ is continuous at $x\in\mathbb{R}$ if and only if $x$ is an rational number. This is not easy to show.

Proof: The function $f$ is defined the following way. If $x$ is irrational number then $f(x)=0$ . If $q$ is a rational number in the form of $\frac{a}{b}$ such that $q$ is its lowest terms (so $a$ and $b$ are relative primes, that is they do not have common prime factors), then let $f(q)=\frac{1}{b}$ . Clearly, $f$ is not continuous at $x$ if $x$ is a rational number. Indeed, there exists a sequence of irrational numbers $\{r_{n}\}^{\infty}_{n=1}$ tending to $x$ and $f(r_{n})=0$ , $f(x)\neq 0$ . On the other hand, a bit surprisingly $f$ is continuous at $x$ if $x$ is an irrational number. Why?

Lemma 4.3.1

Let $a, b, c, d$ be integer numbers such that $c,d<n$ and $\frac{a}{b}\neq\frac{c}{d}$ , then $|\frac{a}{b}-\frac{c}{d}|\geq\frac{1}{n^{2}}$ .

Proof: $|\frac{a}{b}-\frac{c}{d}|=|\frac{ad-bc}{bd}|.$ Now, $|\frac{ad-bc}{bd}|\geq\frac{1}{bd}$ provided that $ad-bc\neq 0$ . Hence the lemma follows. $\square$

Let $x$ be an irrational number and $\varepsilon>0$ . Pick an integer $n>0$ such that $\frac{1}{n}<\varepsilon$ . Then, by the previous lemma in the interval $[x-\frac{1}{4n^{2}},x+\frac{1}{4n^{2}}]$ there is at most one rational number $q$ such that $f(q)\geq 1/n$ . Indeed, if there are two such numbers $\frac{a}{b}$ and $\frac{c}{d}$ then, $b,d\leq n$ and thus $|\frac{a}{b}-\frac{c}{d}|\geq\frac{1}{n^{2}}$ . It is impossible since the length of the interval is $\frac{1}{4n^{2}}$ . Let us chose $\delta>0$ in such a way that there is NO rational number $q$ on the interval $[x-\delta,x+\delta]$ such that $f(q)\geq 1/n$ . This means that if $|x-y|<\delta$ , then $|f(x)-f(y)|<\frac{1}{n}<\varepsilon\,.$ So, the function $f$ is continuous at $x$ . $\square$

Proposition 4.3.4

There exists a function $f:\mathbb{R}\to\mathbb{R}$ such that $f$ is continuous at $x$ if and only if $x$ is not an integer number.

Proof: Let $f$ be defined the following way. $f(x)=exp(x)$ if $x$ is an integer number. Otherwise, let $f(x)=0$ . If $x$ is an integer, then $f(x)\neq 0$ . On the other hand, there is a convergence sequence of non-integers $\{x_{n}\}^{\infty}_{n=1}$ tending to $x$ , so $f(x_{n})=0$ for any $n\geq 1$ , hence $f$ is not continuous at $x$ . Again, if $x$ is not an integer, than there exists a $\delta>0$ such that if $|y-x|<\delta$ , then $y$ is also not an integer, hence $|f(x)-f(y)|=0$ . This shows that $f$ is continuous at $x$ . $\square$

Theorem 4.3.1 (WITHOUT PROOF!!!)

Let $C\subset\mathbb{R}$ be a closed subset. Then there exists a function $f:\mathbb{R}\to\mathbb{R}$ such that set of continuity of $f$ is exactly $C$ . If $g:\mathbb{R}\to\mathbb{R}$ is a function, then its set of continuities $D$ can be a non-closed set (see the proposition above), but it can always be written as $\cup^{\infty}_{n=1}F_{n}$ , where all the $F_{n}$ ’s are closed sets.