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If $A$ and $B$ are subsets of the universal set $S$ then do we know that $A$ intersection (not $B$) is a subset of $A$?

Well, if $S = \{1,2,3,4,5\}$ and $A = B = \{1,2\}$ then not $B = \{3,4,5\}$ and $A$ intersection (not $B$) = $\{\}$ which is not a subset of $A$. So I say that the statement is false? Is this right?

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    $\begingroup$ The empty set is a subset of all sets. $A \cap X \subseteq A$ for all sets $X$ by definition of intersection. $\endgroup$ – Paul Z Jul 22 '14 at 19:57
  • $\begingroup$ If there is an nonempty intersection then those elements will all have to be in $A$, and so $A \cap B^c \subset A$ (assuming $B^c$ is what you mean by "not $B$"). On the other hand, if there is an empty intersection, $A\cap B^c=\emptyset$ then this is okay because $\emptyset$ is a subset of every set, so in particular $A$. $\endgroup$ – Squirtle Jul 22 '14 at 20:17
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Part 0. The comment about $\{\}$ not being a subset of $\{1,2\}$ is incorrect. To check that a set $A$ is a subset of $B$, we need to check, for every $a \in A$, the truthvalue of the statement "$a \in B$". If none of these conditions correspond to false, then we say that $A$ is a subset of $B$.

For example, let us ask: is $A=\{0,1\}$ a subset of $B=\{0,1,2\}$? We need to check the truthvalue of $0 \in \{0,1,2\}$ and $1 \in \{0,1,2\}$. Since none of the conditions being checked correspond to false, hence $A$ is indeed a subset of $B$.

Now let us apply this procedure to work out whether or not $A=\{\}$ is a subset of $B=\{1,2\}.$ We need to check... no conditions. Hence none of the conditions being checked correspond to false, so $A$ is indeed a subset of $B.$

Part 1. The answer to the title question turns out to be Yes.

Proposition. It is a general principle that for all sets $S$ and all subsets $A$ and $B$ of $S$, the following holds.

$$A \cap B \subseteq A$$

Proof. Suppose $x \in A \cap B$. Then ($x \in A \;\& \;x \in B)$. Hence $x \in A$, as required.

Therefore, it is a general principle that for all sets $S$ and all subsets $A$ and $B$ of $S$, the following holds. $$A \cap B^c \subseteq A$$

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