I have a few questions regarding the following notation:

$$ x = (x_1,x_2)\in \{0,1\}^2 $$

Question 1:

Is the following correct?

$\{0,1\}^2$ is the Cartesian product of the 2 sets $\{0,1\}$ and $\{0,1\}$, i.e. \begin{align} \{0,1\}^2 &= \{0,1\} \times \{0,1\} \\ &= \{(0,0),(0,1),(1,0),(1,1)\} \end{align}

Question 2:

With the notation we mean $``$$(x_1,x_2)$ is an element of the set $\{0,1\}^2$$``$, so we can write:

$$ (x_1,x_2)\in \{(0,0),(0,1),(1,0),(1,1)\} $$ So $(x_1,x_2)$ can take the values \begin{align} (x_1,x_2) &= (0,0)\\ (x_1,x_2) &= (0,1)\\ (x_1,x_2) &= (1,0)\\ (x_1,x_2) &= (1,1) \end{align} ?

Question 3:

Does the notation mean that $(x_1,x_2)$ only can assign ONE value of $\{0,1\} \times \{0,1\}$?

I.e. for $(x_1,x_2)$ we have 4 explicit cases: \begin{align} (x_1,x_2) &= (0,0) \\ \text{or} \quad (x_1,x_2) &= (0,1)\\ \text{or} \quad (x_1,x_2) &= (1,0)\\ \text{or} \quad (x_1,x_2) &= (1,1) \end{align}


2 Answers 2


Question 1. Yes, this is exactly the definition of 'square of a set $A$': you consider the cartesian product $A \times A$.

Question 2. Yes, $(x,y)$ belonging to $A^2$ means that it is an element of $A \times A$, so in your case $(x_1, x_2)$ is one of the elements of $\lbrace 0, 1 \rbrace ^2 $.

Question 3. You can have $(x_1,x_2)=(0,0)$, for example, so you have only one of the possible values. This is because $(x_1,x_2)$ is one element of a set.


Yes, yes and yes.

In general, the notation $A^n$, for a set $A$ and natural number $n$, means $$ \underbrace{A \times \ldots \times A}_{n \text{ times}}. $$ So $(x_1, \ldots, x_n) \in A^n$ means that each $x_i$, for $1 \leq i \leq n$, is an element of $A$. Thus this is a tuple of $n$ elements of $A$ (allowing duplicates).


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