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Please do not provide a full answer for this.

Let $2^{S} = \{f : S \rightarrow \{0, 1\}\}$. For $A \subseteq S$, define $\chi_{A}\in2^{S}$ by $$\chi_{A}(s) = \begin{cases} 0 & \text{if } s \notin A \\ 1 & \text{if } s \in A \end{cases}. $$ Show that $\mu : P(S)\rightarrow2^{S}$ given by $\mu(A)=\chi_{A}$ is a bijection.

I know that the standard procedure for showing that a function is bijective is to show that it is both injective and surjective, and the "standard procedures" for those as well. It's just that I don't really know where to start with this.

Seh-kai Liao
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    "I know that the standard procedure for showing that a function is bijective is to show that it is both injective and surjective" ... this is only the standard praised by people who don't understand what bijectivity means, namely invertibility. Better construct a map $\nu : 2^S \to P(S)$ and show that $\mu \nu = \mathrm{id}$ and $\nu \mu = \mathrm{id}$. This is quite easy and more efficient than showing injectivity and surjectivity. – Martin Brandenburg Sep 10 '14 at 01:29
  • @MartinBrandenburg I swear,.. I red your comment after placing my answer. – drhab Sep 10 '14 at 12:53
  • Almost a duplicate: http://math.stackexchange.com/questions/296101/an-explicit-bijection-between-the-power-set-mathcal-p-left-mathbbn-right and http://math.stackexchange.com/questions/41006/how-to-show-equinumerosity-of-the-powerset-of-a-and-the-set-of-functions-from – Martin Sleziak Sep 10 '14 at 13:49

3 Answers3

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For $f\in2^{S}$ define $S_{f}:=\left\{ s\in S\mid f\left(s\right)=1\right\} \subseteq S$ and prove that $\chi_{S_{f}}=f$ and $S_{\chi_A}=A$.

If function $\nu:2^{S}\rightarrow\wp\left(S\right)$ is prescribed by $f\mapsto S_{f}$ then $\nu\circ\mu$ and $\mu\circ\nu$ are identities on $\wp\left(S\right)$ and $2^{S}$ respectively.

This means that both are bijective and eachothers inverse.

drhab
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Injective: show that if $A$ and $B$ are distinct subsets of $S$ (i.e., $A,B \in P(S)$), then $\chi_A$ is different from $\chi_B$.

Surjective: show that any function $f \in 2^S$ can be expressed as $\chi_A$ for some $A$ (that is, construct $A$ so that $\chi_A=f$).

angryavian
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For surjectivity, you should try to show that given any function $\chi\in2^S$ that that there exists some subset $A\subseteq S$ such that $\chi_A=\chi$. Hint: look at the inverse image $\chi^{-1}(1)$.

For injectivity, you can just prove that $A\neq B\implies \chi_A\neq \chi_B$. Hint: one of $A,B$ contains an element that the other doesn't, use this to show that $\chi_A$ and $\chi_B$ disagree for some argument in $S$.

Peter Huxford
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