# $\{0,1\}^n$ and $[0,1]^n$ notations

Does $\{0,1\}^n$ mean a $n$-length vector consisting of $0$s and/or $1$s?

Does $[0,1]^n$ ($(0,1)^n$) mean a $n$-length vector consisting of any number between $0$ and $1$ inclusive (exclusive)?

As a related question, is there a reference web page for all such definitions/notations? Or do we just need to take note of them individually as we learn. Thanks.

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The interpretations you made are the first ones I would think of. But there are too few symbols, too much mathematics. If the writer intends something else, (s)he would have said so. Even if the "natural" interpretation is the intended one, it is useful to remind the reader. – André Nicolas Mar 18 '12 at 22:29
I only know the first one. $\{ 0, 1 \}$ is the binary set, sometimes denoted in as $\mathbb{B}.$ So, yes, $\{ 0, 1\}^{n} = \mathbb{B}^{n};$ understood as $n$-vectors in $\mathbb{B}.$ – user2468 Mar 18 '12 at 22:38
It might help to know what a cartesian product is ... – Neal Mar 18 '12 at 23:12

The notation $\{0,1\}^n$ refers to the space of all $n$-length vectors consisting of $0$s and $1$s, while the notation $[0,1]^n$ ($(0,1)^n$) refers to the space of all $n$-length vectors consisting of real numbers between $0$ and $1$ inclusive (exclusive).

Edit: I often find wikipedia's list of mathematical symbols useful for looking up the meaning of symbols, although I'm not sure it would help with this question.

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+1, but maybe it would be better to call $\left\{0,1{}\right\}^n$ a set of (ordered) n-tuples consisting of 0s and 1s. It certainly can be a space, because it has several structures imposed on it, e.g. order, Hamming distance, etc. But the "n-tuple" wording is slightly more general, I think... – ivt Mar 18 '12 at 23:19
@user825089 I have never seen tuple used to denote an set of elements without order. Could you provide an example? – Alex Becker Mar 18 '12 at 23:20
I haven't either, but apparently the term exists as a natural extension of unordered pair: en.wikipedia.org/wiki/Unordered_pair (last sentence). – ivt Mar 18 '12 at 23:28

The idea is simple… The abstract set of Topology

$$\displaystyle{\prod_{i\in I}X_i=\left\{x:I\rightarrow\cup_{i\in I}X_i \vert x(i)=x_i\in X_i,\;\forall i\in I \right\}}$$

where $x$ are continuous functions. (Also, you can see the continuous function as equivalence relation with $n$-length vector)

Examples:

1. If $X_i=\{0,1\}, \forall i\in I$ then $$\displaystyle{\prod_{i\in I}X_i=\{0,1\}^I}$$ $$x=(x_i)_{i\in I}\in\{0,1\}^I$$

2. If $X_i=[0,1], \forall i\in I$ then $$\displaystyle{\prod_{i\in I}X_i=[0,1]^I}$$

3. Important case is the Cantor's set.

P.D.: Excuse my English, please.

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Not that "simple" for a simple idea. What you're saying is right, but you're... a little intense for OP. I think. =P – Patrick Da Silva Mar 18 '12 at 23:13