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Our math prof gave us this and we are not sure what to make of the notation, can someone give us a hand?

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@Moron: Why are you so sure that those are the two things that are causing problem? There could also be an issue with $2^A$ for that matter... –  Arturo Magidin Nov 10 '10 at 4:50
    
@Arturo: I am not sure. BinaryBro can always edit the title if there is something missing. The earlier title was too generic. –  Aryabhata Nov 10 '10 at 4:58
    
@Moron: fair enough... –  Arturo Magidin Nov 10 '10 at 4:59
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2 Answers 2

$\mathcal{P}(A)$ is the power set of $A$; it is a set whose elements are precisely the subsets of $A$. For example, if $A=\{1,2,3\}$, then $\mathcal{P}(A) = \{\emptyset, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}\}$.

Given any two sets $A$ and $B$, $A^B$ represents the set of all functions from $B$ to $A$. So $2^A$, with "$2$" meaning the set $\{0,1\}$ means the set of all functions $f\colon A\to \{0,1\}$; all functions whose domain is $A$, and which take only the values $0$ and $1$.

If $A$ is a set and $S$ is a subset, then $\chi_S$ is the characteristic function of $S$. It is a function $\chi_S\colon A\to \{0,1\}$ defined by: $$\chi_S(a) = \left\{\begin{array}{ll} 0 &\mbox{if $a\not\in S$,}\\ 1 & \mbox{if $a\in S$.} \end{array}\right.$$

By "$\chi_{-}$" he means the function that takes each subset $S$ to the characteristic function $\chi_S$.

There is no new notation in part (b).

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$\mathcal{P}(A)$ is the power set of $A$, i.e., the set of subsets of $A$, and $2^A$ is the set of functions from $A$ to $\{0,1\}$. The characteristic function $\chi_S$ of a subset $S$ of $A$ is the function $\chi_S:A\to\{0,1\}$ defined by $\chi_S(x)=1$ if $x$ is in $S$ and $\chi_S(x)=0$ if $x$ is not in $S$.

In general it is a good idea to ask your professor to clarify notation on the homework if it is not easy to find in the notes or text.

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Thanks, bro. This helps. –  BinaryBro Nov 10 '10 at 2:34
    
Hmm... came up just as I was typing. (-: –  Arturo Magidin Nov 10 '10 at 2:41
    
If I had known you were working on a more informative answer I probably wouldn't have posted. I guess I gave the short attention span version. :) –  Jonas Meyer Nov 10 '10 at 2:46
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Nah; your answer is perfectly fine. I'm just (in)famous for long writings. Just ask my old Algebraic Geometry TA. "Tree-killer", they used to call me. –  Arturo Magidin Nov 10 '10 at 2:48
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