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Let $\Omega$ be the universal set (which contains all objects of interest) and let its complement be denoted by $\Omega^c$ . The book I'm reading states that $\Omega^c = \phi$ ; does that mean $\phi \not\subset \Omega$?

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12  
$\in\ne\subset$ –  Andres Caicedo Dec 21 '12 at 22:58
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No, the empty set is a subset of every set (every element of the empty set is contained in any other set). –  copper.hat Dec 21 '12 at 23:05
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@Jacob : No, and it is not precisely because the empty set is empty. Being in both $\Omega$ and $\Omega^c$ implies being in $\phi$. Nothing is wrong with that. –  Patrick Da Silva Dec 21 '12 at 23:07
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It would be if the empty set had any elements. –  Michael Albanese Dec 21 '12 at 23:07
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@Jacob : The complement of a set $A \subseteq \Omega$ is defined as the elements of $\Omega$ that are not in $A$. So of course, $A \cap A^c = \varnothing$. This is also true if $A = \Omega$. (Note that being a subset of $A$ and $A^c$ implies being a subset of $A \cap A^c$, so that this answers your last comment.) –  Patrick Da Silva Dec 21 '12 at 23:13

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Let's make a community wiki answer based on the above comments. Given a subset $A \subset \Omega$, the relative complement $A^c$ in $\Omega$ is defined as the set of elements of $\Omega$ that are not elements of $A$. So in the case that $A = \Omega$, the relative complement of $A$ in $\Omega$ is equal to $\emptyset$. That is, for every set $x$ we have $x \notin A^c$. In particular we have $\emptyset \notin A^c$. However, this is probably the "wrong" question to ask, because $\emptyset$ might not be an element of $\Omega$ at all. Notice that the empty set is a subset of every set, so in particular $\emptyset \subset \Omega$. (This is because to check whether $X \subset Y$ you take an arbitrary element of $X$ and check that it is also in $Y$. If $X$ is empty, then there is nothing to check.)

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protected by Alexander Gruber Jun 12 '13 at 2:13

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