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Is the Cantor function bijective from $[0,1]$ to Cantor set $K$? As $K$ is uncountable I think cardinality of $K$ must be $\mathfrak c$ as $K$ is a subset of $[0,1]$. But I am surprised whether there is any connection between cardinality and measure of a set. But in particular I am asking cardinality of $K$ and if it is $\mathfrak c$ give me a bijection with $K$ and $\mathbb R$.

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The Cantor function that I know is not a bijection. If $\frac 13 \lt x \lt \frac 23, f(x)=\frac 12$ It is true that the cardinality of $K$ is $\mathfrak c$. In the reals, all sets that have positive measure have cardinality $\mathfrak c$, but there are sets ($K$ among them) that have cardinality $\mathfrak c$ and measure $0$

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  • $\begingroup$ ok, Cantor function is not bijective. But give me some proof that card K = c $\endgroup$ – EuReka Dec 20 '13 at 16:49
  • $\begingroup$ Give a proof that card(K)=c $\endgroup$ – EuReka Dec 20 '13 at 16:51
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    $\begingroup$ Have you seen that $K$ consists of all base $3$ expansions that have only $0$'s and $2$'s? That is uncountable by the usual Cantor diagonal argument. You can also change the $2$'s to $1$s and read them as binary to get an (up to a countable set) bijection to $(0,1)$ $\endgroup$ – Ross Millikan Dec 20 '13 at 16:52
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    $\begingroup$ The Cantor function is not bijective, but it does map $K$ onto $[0,1]$, showing the cardinal of $K$ is at least the cardinal of $[0,1]$. $\endgroup$ – GEdgar Dec 20 '13 at 17:17
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    $\begingroup$ Then what maps to $0.2?$ in binary $0.1=0.011111\dots $ You can use either, but not both. One violates surjectivity. $\endgroup$ – Ross Millikan Dec 20 '13 at 21:27

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