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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.

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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.

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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?

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  • $\begingroup$ 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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