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Does there exist a function $f : \mathbb{R} \rightarrow \mathbb{R}$ that is strictly increasing and discontinuous everywhere?

My line of thought (possibly incorrect): I know there are increasing functions such as $f(x) = x$, and there are everywhere-discontinuous functions such as the Dirichlet function. I also know that when there is a discontinuity at a point $c$, there is a finite gap $\epsilon$ such that there are points $d$ arbitrarily close to $c$ such that $|f(d) - f(c)| > \epsilon$. This is where my thinking gets unclear - does it make sense to have a "gap" at every real number?

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    @Jonas Meyer: Thanks. I blundered. I meant, as you said, "except on a set of measure 0". Tnanks again. – 2012-01-02

2 Answers 2

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There is no such function. Suppose that $f:\mathbb{R}\to\mathbb{R}$ is strictly increasing. For each $a\in\mathbb{R}$ let $f^-(a) =$ $\lim\limits_{x\to a^-}f(x)$ and $f^+(a) = \lim\limits_{x\to a^+}f(x)$. Then $f$ is discontinuous at $a$ if and only if $f^-(a) < f^+(a)$. Let $D = \{a\in\mathbb{R}:f\text{ is not continuous at }a\}$, and for each $a\in D$ let $q_a$ be a rational number in the non-empty open interval $I_a = (f^-(a),f^+(a))$.

It’s not hard to check that if $a,b \in D$ with $a, then $f^+(a) \le f^-(b)$, so the intervals $I_a$ are pairwise disjoint. This implies that the rational numbers $q_a$ are all distinct. (If you want to be fancy, the function from $D$ to $\mathbb{Q}$ that sends $a$ to $q_a$ is injective.) But there are only countably many rational numbers, so the set $D$ must be countable. In other words, the function $f$ can have at most countably many points of discontinuity. And of course $\mathbb{R}$ is uncountable, so $f$ cannot be discontinuous everywhere.

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    @Anoldmaninthesea. if you did that, your function must be undefined in an uncountable set of the domain and it has to be defined everywhere on the domain. and if you define it such that the limit does not exist at the irrationals, it can not be increasing or the function is not well defined the domain... – 2017-08-23
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Another way to see that it is impossible to have uncountably many of these "gaps" is to first consider the restriction of $f$ to a bounded interval $(a,b)$, where it is bounded. If $\varepsilon>0$, the number of points in $(a,b)$ where there is a gap of size greater than $\varepsilon$ is finite, less than $\frac{1}{\varepsilon}(f(b)-f(a))$. Taking countably many $\varepsilon$s going to zero allows you to conclude that there are only countably many discontinuities in $(a,b)$. Taking countably many $(a,b)$s whose union is $\mathbb R$ allows you to finish.

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    Generalizing this, if $f:(a,b)\to\mathbb R$ is monotone, then for each $c$ and $d$ with a, $f$ is bounded on $(c,d)$, with $f(c)\leq f(x)\leq f(d)$ for all $x\in(c,d)$. Note that in my answer, I was specific about how boundedness of $f$ on the bounded interval comes in where I mentioned the bound on the number of gaps of a given size. Your example is irrelevant because $\frac{1}{1-x}$ has no monotone extension to $\mathbb R$. – 2011-08-13