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If $S$ is a subset, is $S+\emptyset$ defined and equals to $S$? Or is it just gibberish? Thanks again. 

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Assuming that $A+B = \{ a + b \,:\, a \in A, b \in B \}$, we have $S + \{ \mathbf 0 \} = S$ and $S + \emptyset = \emptyset$. ($\mathbf 0$ is the additive identity, aka the zero vector.) –  Srivatsan Sep 10 '11 at 23:35

2 Answers 2

$A+B$ means $\{a+b|a \in A, b \in B\}$. That is, it is the set of all points which are the sum of a point in $A$ and a point in $B$. The empty set $\emptyset$ has no points in it, so there is nothing that is the sum of a point in $S$ and a point in $\emptyset$. Thus for all $S$, $S + \emptyset=\emptyset$.

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This can be true only if there exists an algebraic operation $+$ between elements of $A$ and $B$. Otherwise, how do you define $a+b$? –  Manos Sep 10 '11 at 23:30
    
@Manos: Since the OP tagged the question linear-algebra, it seemed clear we're talking about Minkowski sum. –  Chris Eagle Sep 10 '11 at 23:31
    
Indeed. Thanks. –  Dave Sep 10 '11 at 23:39
    
@Chris: If it is clear, then why was it not clear to me? But thanks for the clarification. –  Manos Sep 11 '11 at 0:45

Both $S$ and the empty set are sets. If we define an operation that admits two sets and maps them to some other mathematical entity (e.g. another set), then we could apply this operation to $S$ and the empty set. What we will obtain as a result of the operation depends on how we have defined our operation. For example we could interpret the $+$ as union, or intersection of sets. In the first case we would get $S$, in the second case we would get the empty set.

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