# Removing elements from an empty set

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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What does "removing elements" mean, and how are you going to try to remove them? Certainly any subset of the empty set is itself empty. –  Arturo Magidin Jul 21 '11 at 16:26
For example, $(e - A)$ can be interpreted as removing all elements of set $A$ from $e$. But if $e = \emptyset$, would $(e - A) = \emptyset$ ? –  Student Jul 21 '11 at 16:31
Yes, it would be empty, because $A-B$ is always a subset of $A$. –  Arturo Magidin Jul 21 '11 at 16:33
@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 $e$. –  joriki Jul 21 '11 at 16:35

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 $\emptyset\setminus\{1,2\}=\emptyset$.
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 $B$.