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I am having difficulty answering the following question on "Notes on Set Theory" by Moschovakis, 1st edition:

Prove that the set $K$ of all monotone real functions on the closed interval $[0,1]$ have the same cardinality with $\mathbb R$.

We can inject $\mathbb R$ into $K$ by sending any real number to its constant function, e.g. $5/2$ gets mapped to the constant function $5/2$ on $[0,1]$. By Schroder-Berstein theorem, it suffices to provide an injection from $K$ to $\mathbb R$. As the set of constant functions in $K$ have cardinality of $\mathbb R$, strictly decreasing function in $K$ (by which I mean that such a function $f$ has some $a > b$ st. $f(a) <f(b)$, i.e. $f$ need not be strictly decreasing in the whole interval) is in bijective correspondence with strictly increasing function in $K$, and $|\mathbb R^3| = |\mathbb R|$, it suffices to show that the set of strictly increasing function in $K$ injects to $\mathbb R$. However, I am not sure how I can provide such an injection.

Andrés E. Caicedo
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Squirrel-Power
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1 Answers1

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The set of discontinuities of an increasing function is at most countable.

Therefore $f \in K$ is completely described by its set of discontinuities $D$ and its values on $\mathbb{Q} \cap [0,1]$.

Indeed, if $f$ is continuous at $x$, then $f(x) = \lim_{n\to\infty} f(q_n)$ where $(q_n)_n$ is some sequence of rationals converging to $x$. If $f$ is discontinuous at $x$, then the value $f(x)$ is already given.

Therefore we have a bijection of $K$ with

$$\{(D, (x_n)_n) : D \subseteq [0,1]\text{ at most countable}, x_n \in [0,1], \forall n\in\mathbb{N}\}$$

so $$|K| = c^{\aleph_0} \cdot c^{\aleph_0} = c$$

mechanodroid
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