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I'm trying to find a bijection between the Cantor set and $[0, 1]$. Is this a valid proof? If not, am I on the right track?

We prove there exists a bijection between the Cantor set $\mathcal{C} = \bigcap_{n = 0}^{\infty} C_n$ (where $C_0 = [0, 1]$, $C_1 = [0, \frac{1}{3}] \cup [\frac{2}{3}, 1]$, and so on) and $[0, 1]$ via induction on $n$. For our base case $n = 0$, we use the trivial bijection. Now, assume there exists a bijection from $C_k$ to $[0, 1]$ for some $k \in \mathbb{N}$. We want to show the same for $C_{k + 1}$; equivalently, we can find a bijection from $C_{k + 1}$ to $C_k$. Since $C_k$ has $2^k$ connected subsets, whereas $C_{k + 1}$ has $2^{k + 1}$ connected subsets, we can simply create a map that appropriately scales and translates each of the $2^{k + 1}$ connected components of $C_{k + 1}$ to those of $C_k$, hence we have a bijection.

Since we can find a bijection for any $n \in \mathbb{Z}$, and since $\mathcal{C} = \lim_{n \to \infty} C_n$, we can find a bijection from $\mathcal{C}$ to $[0, 1]$

In particular, I'm not sure if the last sentence of my proof is valid.

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    $\begingroup$ The last sentence is definitely invalid. To see that, just recall that you can find a bijection between every set $(0,\frac{1}{n})$ and $(0,1)$, but $\bigcap_{n\in\mathbb N}(0,\frac{1}{n})=\emptyset$. $\endgroup$ – user491874 Jan 21 '18 at 9:51
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    $\begingroup$ Also, "appropriately scaled and translated" is imprecise - try to formalise it and you will see you actually have trouble with interval endings: e.g. in the map $[0,\frac{1}{3}]\cup[\frac{2}{3},1]\to[0,1]$, how do you ensure that $\frac{1}{3}$ and $\frac{2}{3}$ are mapped into different points? $\endgroup$ – user491874 Jan 21 '18 at 9:55
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    $\begingroup$ See this. It does not give you a bijection but a surjection, however it's enough to prove there is a bijection. $\endgroup$ – Jean-Claude Arbaut Jan 21 '18 at 9:56
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No, the last sentence is not valid. For each $n\in\mathbb N$, there is a bijection between $\left[0,\frac1n\right]$ and $\mathbb R$. However, $\bigcap_{n\in\mathbb N}\left[0,\frac1n\right]=\{0\}$ and there is no bijection between $\{0\}$ and $\mathbb R$.

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The statement:

"Since we can find a bijection for any $n∈Z$, and since $ \mathcal{C} = \lim_{n \to \infty} C_n $, we can find a bijection from $C$ to $[0,1]"$

is not true.

You may find sets such as $(-1/n,1/n)$ where each set in one- to- one correspondence with the real line but the intersection contains only one point.

You have to approach your problem with a different strategy such as using binary sequences.

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