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I am currently trying to understand a proof from here that all perfect sets have the same cardinality as $\mathbb{R}$.

So given some perfect set $P \subseteq \mathbb{R}$, the identity mapping $\text{id}_{P}: P \rightarrow \mathbb{R}$ given by $x \mapsto x$ is an injection and so now the goal is to create an injection from $\mathbb{R}$ to $P$ and then invoke the Schroder-Bernstein theorem.

The method outlined in the link is to construct an injection from the set of infinite binary sequences to $P$. They state that we can associate to each infinite binary sequence $\xi = \xi_{0}\xi_{1}\xi_{2}\cdots$ a real number in the interval $[0,1]$ via the mapping $$ \xi \mapsto \sum_{i \geq 0}\frac{\xi_{i}}{2^{i +1}} .$$ They then state that the cardinality of $P$ is at least as large as the cardinality $[0,1]$. This is the part I am having trouble with, i.e.

Why is the cardinality of $P$ at least as large as the cardinality of $[0,1]$?

To me it seems that we have constructed an injection from the set of infinite binary sequences (which has the same cardinality as $\mathbb{R}$) into $[0,1]$ and not into $P$.

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  • $\begingroup$ The linked proof looks wrong to me. They're saying there's a surjective mapping from the set of all infinite sequences of $0$s and $1$ to the set $P$, and from that they deduce that $P$ has at east as many members as does the set of such sequences. That doesn't make sense. The existence of a surjective mapping from $A$ to $B$ does not mean $B$ has at least as many members as $A$. Furthermore the mapping from the set of all sequences $\xi = \xi_1,\xi_2,\xi_3,\ldots$ is defined as $\xi\mapsto \sum_{i\ge0} \xi_i 2^{-i-1}$. That maps onto the interval $[0,1]$; there's no reason to$\,\ldots$ $\endgroup$ – Michael Hardy Apr 5 '16 at 20:14
  • $\begingroup$ $\ldots\,$think that it maps into the set $P$, which was an arbitrary perfect set. The proof makes no sense to me at all. $\qquad$ $\endgroup$ – Michael Hardy Apr 5 '16 at 20:14
  • $\begingroup$ @MichaelHardy Okay, I thought that the proof was suspect as well. $\endgroup$ – Oiler Apr 5 '16 at 22:42
  • $\begingroup$ You can find a correct proof here. $\endgroup$ – Brian M. Scott Apr 5 '16 at 23:32

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