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I have been reading "Set theory" of T. Jech. I saw a proof of "Every perfect subset (in $\mathbb{R}$) has cardinality $\mathfrak{c}$". Here is this proof:

Proof: Given a perfect set $P$, we want to find a one-to-one function $F$ from $\{0,1\}^\omega$ into $P$. Let $S$ be a set of all finite sequence of 0's and 1's. By induction on the length of $s\in S$ one can find closed intervals $I_s$ s.t. for each $n$ and all $s\in S$ of length $n$,

(i) $I_s\cap P$ is perfect,

(ii) the diameter of $I_s$ is $\le 1/n$,

(iii) $I_{s\frown 0}\subset I_s$, $I_{s\frown 1}\subset I_s$ and $I_{s\frown 0}\cap I_{s\frown 1}=\varnothing$.

(The rest is omitted.)

But I do not understand this proof. Especially, I do not understand about existence of $I_s$. I think $I_s$ can be taken by without the axiom of choice, but I cannot find out this method. Thanks for any help.

Community
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Hanul Jeon
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  • If $P$ has nonempty interior, the conclusion is trivial. So we may assume that $P$ is nowhere dense, or in other words, its interior is empty. This will help you construct such a sequence. Just imagine a Cantor set as a typical example that would help you understand what is going on this proof. In fact, this proof shows that every perfect subset of $\Bbb{R}$ is in some sense Cantor-set-like. – Sangchul Lee Apr 15 '13 at 09:43

1 Answers1

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As Jech says, the construction of the sets $I_s$ is by induction on the length of $s$. So suppose we already have $I_s$ for a certain $s\in S$ of length $n$, and we want to find suitable $I_{s^\frown0}$ and $I_{s^\frown1}$. Since $I_s\cap P$ is perfect, it contains two points $a$ and $b$ that are not endpoints of $I_s$. Let $r$ be a positive number smaller than all of the following: $1/(n+1)$, the distance between $a$ and $b$, the distance from $a$ to the nearer endpoint of $I_s$, and the distance from $b$ to the nearer endpoint of $I_s$. As a first guess, let $I_{s^\frown0}$ and $I_{s^\frown1}$ be the closed intervals of length $r$ centered at $a$ and at $b$, respectively. Conditions (ii) and (iii) are satisfied because of our choice of $r$. Condition (i) might fail, but only if an endpoint of $I_{s^\frown0}$ or $I_{s^\frown1}$ is isolated in $I_{s^\frown0}\cap P$ or $I_{s^\frown1}\cap P$; in that case, shrink the two intervals slightly to achieve (i); the shrinking only helps (ii) and (iii).

Andreas Blass
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    Note that all of this can be done rather explicitly, so no amount of choice is needed for this proof. On the surface, we are choosing at each stage the points $a$ and $b$, but this can be done explicitly since $I_s\cap P$ is closed, so, for any $x$, there is a point of $I_s\cap P$ closest to $x$. – Andrés E. Caicedo May 08 '13 at 01:02