Why is $\displaystyle B \cap \{ \emptyset \} = \emptyset$? Shouldn't it be $\displaystyle B \cap \{ \emptyset \} = \{\emptyset \}$?
Last edited by mr fantastic; Sep 29th 2011 at 10:31 PM. Reason: Fixed latex.
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Originally Posted by Jskid Why is $\displaystyle B \cap \{ \emptyset \} = \emptyset$? Shouldn't it be $\displaystyle B \cap \{ \emptyset \} = \{\emptyset \}$? It can be either depending on whether $\displaystyle \emptyset\in B$.
Originally Posted by Jskid Why is $\displaystyle B \cap \{ \emptyset \} = \emptyset$? $\displaystyle B \cap \{ \emptyset \} = \emptyset$ is NOT a theorem. Who said it is? Originally Posted by Jskid Shouldn't it be $\displaystyle B \cap \{ \emptyset \} = \{\emptyset \}$? As emakarov alluded $\displaystyle B \cap \{ \emptyset \} = \{\emptyset \}$ if and only if $\displaystyle \emptyset\in B$
First page of section 2.2 in Discrete Mathematics for Logic and Foundations.
are you sure about the name of the book. amazon doesn't come up with any book with that name
Originally Posted by Jskid First page of section 2.2 in Discrete Mathematics for Logic and Foundations. Then, if you're quoting the book correctly, taking you to mean that the book claims it's a theorem of ordinary set theory, then the book is in clear error.
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