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So if I asked: "Is there an explicit function / no-axiom-of-choice construction $f:[0,1] \to [0,1]$ so that COUNTABLY many disjoint subsets of $[0,1]$ map onto $[0,1]$?" The answer would be yes, because we could construct disjoint closed subintervals of length $1/4,1/8,1/16,\ldots$ and then use an affine map to map each subinterval onto $[0,1]$.

But what if we want UNCOUNTABLY many disjoint subsets of $[0,1]$ to map onto $[0,1]$? Is some form of choice required? Or can we come up with an explicit function / no-axiom-of-choice construction?

user2566092
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3 Answers3

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Let $(X(t),Y(t)), t=0\ldots1$ be a space-filling curve, a continuous map from $[0,1]$ onto $[0,1]\times [0,1]$. Many explicit examples are known, e.g. Peano's. The sets $S(y) = \{t \in [0,1]: Y(t) = y\}$ are disjoint, and $X$ maps each onto $[0,1]$.

Robert Israel
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    It's also worth mentioning that each of the sets $S(y)$ is closed. I think people sometimes assume that such sets would have to be pathological, but in this example they're really not. – Nate Eldredge Oct 01 '15 at 17:05
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There is an easy surjective map from the middle-thirds Cantor set $C \subseteq [0,1]$ to the unit interval: we take a base-3 expansion of $x \in C$ with no $1$s, then replace all the $2$s in the expansion with $1$s, and regard the result as a base-2 expansion of a real $f(x)$ in $[0,1]$.

Now $C$ is homeomorphic to $C \times C$ by an explicit map, so this gives us a map from $[0,1]$ to itself that sends continuum many disjoint subsets of $[0,1]$ onto $[0,1]$: let $C \times C \ni (a,b) \mapsto f(b) \in [0,1]$. (For concreteness, let the map be constant on points that are not in the Cantor set).

Carl Mummert
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This answer gives an explicit bijection between $(0,1]$ and $(0,1]\times(0,1]$ and another between $(0,1]$ and $[0,1]$; put them together properly, and you have an explicit bijection

$$h:[0,1]\to[0,1]\times[0,1]\;.$$

For $y\in[0,1]$ let

$$A_y=\big\{x\in[0,1]:f(x)\in\{y\}\times[0,1]\big\}\;,$$

let

$$\pi:[0,1]\times[0,1]\to[0,1]:\langle y,z\rangle\mapsto z\;,$$

and define $f=\pi\circ h$. Then $f:[0,1]\to[0,1]$, $f[A_x]=[0,1]$ for each $x\in[0,1]$, and the family $\{A_x:x\in[0,1]\}$ is pairwise disjoint.

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Brian M. Scott
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