I was given the following task

Let $\Omega$ - be some set (finire or infinite)
Assign to each subset A $\subset\Omega$ function $\chi_A : \Omega \to {(0,1)}$
Where $\chi_A$ is defined as $\chi_A(x) = 1 $ if $x \in A$ and $\chi_A(x) = 0 $ if $x \notin A$

I need to prove that

$A \mapsto \chi_A$ defines bijection between $\cal P(\Omega)$ (all the subsets of set $\Omega$) and the set of functions ${(0,1)}^{\omega}$

In this problem, I do not understand the following :

  1. What does it mean that some function defines a bijection between set A and set B.
  2. What is the set of functions ${(0,1)}^{\omega}$
  • $\begingroup$ What is $\beta(\Omega)$? $\endgroup$ – J.-E. Pin Sep 16 at 6:19
  • $\begingroup$ $\beta(\Omega)$ all the subsets of set $\Omega$ $\endgroup$ – Rustem Sadykov Sep 16 at 6:22
  • $\begingroup$ The standard notation for the set of all subsets of $\Omega$ is ${\cal P}(\Omega)$. $\endgroup$ – J.-E. Pin Sep 16 at 6:25

There are several notation issues in your question. The right statement of the question should be

Show that the function $\chi:A \to \chi_A$ defines a bijection from ${\cal P}(\Omega)$ to $\{0,1\}^\Omega$.

The set $\{0,1\}^\Omega$ is the set of all functions from $\Omega$ to $\{0,1\}$. In particular, for each subset $A$ of $\Omega$, $\chi_A$ belongs to $\{0,1\}^\Omega$.

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