Let $f$ be a strictly increasing function (that is, $f(b) > f(a)$ if $b > a$). Show that $f$ is continuous at at some point.

Hint: Use the fact that uncountably many positive real numbers can not have a finite sum.

This leads me to the following idea. Let $J = \mathbb{R} \setminus f(\mathbb{R})$.

If $J = \varnothing$, then $f$ is surjective. Consider any $x$, and fix $\varepsilon > 0.$

Take $a$ and $b$ such that $f(a) = f(x) - \varepsilon$ and $f(b) = f(x) + \varepsilon.$ Since the function is strictly increasing, we must have $x \in (a,b)$ and for any other $x'$ in the interval we have:

$a < x' < b \implies f(x) < f(x') < f(b) \implies |f(x') - f(x)| < \varepsilon.$

Then choose $\delta = \min \{|x-a|, |x-b| \}$ to ensure $a \leq x - \delta < u < x+\delta \leq b.$ Thus if $|x-u| < \delta,$ we have $|f(x) - f(u)| < \varepsilon$ as desired.

The case that $J$ is countable can be dealt with quite similarly. This leaves the case where $J$ is uncountable, and without loss of generality we can consider $J \cap (0,\infty)$ uncountable as well by considering the function $g$ where $g(x) = f(x) + a$.

This seems to be the point where the hint might come in to play? Any further hints would be greatly appreciated.


Notice that by monotonicity, all the intervals of the form $(\lim\limits_{x\to x_0^-}f(x),f(x_0))$ and $(f(x_0),\lim\limits_{x\to x_0^+}f(x))$ are disjoint for all $x\in \mathbb R$.

Suppose that $f$ is discontinuous at $x_0$, this means that $\lim\limits_{ x_\to x_o^+}f(x)$ or $\lim\limits_{x\to x_0^-}f(x)$ is different from $x_0$. and so one of the two open open intervals is non-empty.

This implies that $f$ must have a numerable number of discontinuities, since each non-empty open interval contains rational points.


Suppose $f$ is such a function. Let $a,b\in \Bbb{R}$ with $a < b$. Given $x\in (a,b)$, define $$d_x = \inf_{y > x} (f(y)-f(x))$$ The fact that $f$ is increasing means that the set of all $f(y)-f(x)$ with $y>x$ is bounded below by $0$, guaranteeing that this $\inf$ exists. Now choose points $x_0, x_1, ..., x_n$ such that $$a=x_0 < x_1 < ... < x_{n-1} < x_n = b$$ then $$f(b) - f(a) \geq (f(x_n) - f(x_{n-1}))+...+(f(x_1)-f(x_0)) \geq d_{x_{n-1}}+...+d_{x_0}$$ Since the values $n$ and $x_1, ..., x_{n-1}$ are arbitrary, it follows that $$f(b)-f(a) \geq \sum_{x\in [a,b)}d_x$$ Can you finish the proof?

  • $\begingroup$ I'm a little confused about the ending - I can believe the last line follows for a countable sum but don't really understand how the same logic would carry forward for an uncountable sum. $\endgroup$ – Nitin Dec 30 '16 at 14:46
  • $\begingroup$ The sum of elements of a set of reals $S$ is defined as the supremum of all of the finite sums of elements from $S$. Let $S$ be the set of all $d_x$ where $x\in [a,b)$. Because $f(b)-f(a)$ is greater than or equal to any finite sum of elements of $S$, it is greater than or equal to the supremum of such finite sums, and so $f(b)-f(a)$ is greater than or equal to the infinite sum. $\endgroup$ – florence Dec 30 '16 at 19:56

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