The Cantor Space as $\{0,1\}^{\mathbb{N}}$ and as $[0,1]$.

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The Cantor-Space is defined as the space of all infinite binary sequences, i.e. the space $\{0,1\}^{\mathbb{N}}$. It has a natural metric, $$ d(x,y) = \inf\{ 2^{-|w|} : w \in pref(x) \cap pref(y) \} $$ where $pref(x)$ denotes the set of all prefixes of $x$ (i.e. finite initial segments of an infinite binary sequence/string).

Let $X = \{0,1\}$ and $X^*$ denote all finite sequences over $X$. Then the open, closed and clopen sets are characterised as follows:

1) Open sets are of the form $W \cdot X^{\mathbb{N}}$, where $W \subseteq X^*$.

2) A subset $E \subseteq X^{\mathbb{N}}$ is open and closed (clopen) iff $E = W \cdot X^{\mathbb{N}}$ where $W \subset X^*$ is finite.

3) A subset $F \subseteq X^{\mathbb{N}}$ is closed iff $F = \{ \xi : pref(\xi) \subseteq pref(F) \}$.

Now every $x = x_1 x_2 x_3 \ldots \in X^{\mathbb{N}}$ could be identified with an element from $[0,1]$ by the dyadic interpretation $$ \sum_{i=1}^{\infty} \frac{1}{2^i} x_i $$ and by identifiying sequences of the form $010000...$ and $001111...$ one gets an isomorphism. Now I ask myself how look the images of closed, open and clopen sets under this isomorphism, if $x$ is closed, then the image is also a closed subset of $[0,1]$, but what about the open sets?

EDIT: Corrected definition of $d(x,y)$.

EDIT: Corrected last equation.

EDIT: To clarify point 3).

Lemma: Let $L$ be a subset of $X^{\mathbb{N}}$ and let $u \in X^{\mathbb{N}}$. Then $u \in \overline{L}$ iff every prefix of $u$ is a prefix of a word of $L$.

Proof: We have $u \in \overline{X}$ iff for each $n \ge 0$ there exists $v \in L$ such that $d(u,v) < 2^{-n}$. This amounts to saying that, for each $n \ge 0, u_1\ldots u_n$ is a prefix of a word of $L$. q.e.d.

Note that $d(x,y)$ could also be written: $$ d(x,y) = 2^{-r} \quad where \quad r = \operatorname{min}\{ n | x_n \ne y_n \}. $$