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If I try to remove elements from an empty set, would I get an empty set or would this operation be undefined?

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    @Jon: $(e-A)$ cannot be interpreted as removing all elements of $A$ from $e$, since $e$ might not contain all elements of $A$. Rather, $(e-A)$ is defined as the result of removing all elements of $A$ contained in $e$ from $e$.2011-07-21

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Obviously you cannot actually remove elements from an empty set, so I'm guessing that your question is intended to refer to the operator $\setminus$, as in $\emptyset\setminus\{1,2\}$. If this is so, the question pertains not only to the empty set, but to any case in which the left-hand operand contains elements that the right-hand operand doesn't contain. The operator is defined such that these elements are irrelevant, and only the elements actually contained in the left-hand operand are actually removed; so the operation is well-defined and yields $\emptyset\setminus\{1,2\}=\emptyset$.

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    Thank yo$u$, this makes much more sense!2011-07-21
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A definition of the set $A\setminus B$ (read "A setminus B") is $ A\setminus B=\{x\,|\,x\in A,x\notin B\}. $ In particular, $A\setminus B\subseteq A$. If $A=\emptyset$, this proves that $A\setminus B=\emptyset$ for every set $B$.

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    As @Arturo said.2011-07-21