Difference between revisions of "Test"

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<math>\frac{1}{\displaystyle1+\frac{1}{\displaystyle 1+\sqrt{5}}}</math>
 
<math>\frac{1}{\displaystyle1+\frac{1}{\displaystyle 1+\sqrt{5}}}</math>
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<wikitex>
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Let $Q$ be any finite set, and $\mathcal B=2^Q$ be the collection of the subsets of
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$Q$. Let $f:\mathcal B\rightarrow \mathbb R$ be a function assigning real numbers to
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the subsets of $Q$ and suppose $f$ satisfies the following conditions:
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:(i) $f(A)\ge 0$ for all $A\subseteq Q$, $f(\emptyset)=0$,
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:(ii) $f$ is monotone, i.e. if $A\subseteq B\subseteq Q$ then $f(A)\le f(B)$,
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:(iii) $f$ is submodular, i.e. if $A$ and $B$ are different subsets of $Q$ then
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      $$f(A)+f(B)\ge f(A\cap B) + f(A\cup B).\eqno{(2)}$$
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</wikitex>

Revision as of 14:01, 18 November 2011

<math>\frac{1}{\displaystyle1+\frac{1}{\displaystyle 1+\sqrt{5}}}</math>

<wikitex> Let $Q$ be any finite set, and $\mathcal B=2^Q$ be the collection of the subsets of $Q$. Let $f:\mathcal B\rightarrow \mathbb R$ be a function assigning real numbers to the subsets of $Q$ and suppose $f$ satisfies the following conditions:

(i) $f(A)\ge 0$ for all $A\subseteq Q$, $f(\emptyset)=0$,
(ii) $f$ is monotone, i.e. if $A\subseteq B\subseteq Q$ then $f(A)\le f(B)$,
(iii) $f$ is submodular, i.e. if $A$ and $B$ are different subsets of $Q$ then
     $$f(A)+f(B)\ge f(A\cap B) + f(A\cup B).\eqno{(2)}$$

</wikitex>