A set \(A\) is a subset of a set \(B\text{,}\) denoted \(A \subseteq B\) , if every element of \(A\) is also an element of \(B\text{.}\) that is , \(\forall x (x \in A \implies x \in B)\text{.}\)
Let \(A = \set{a, b, c}\text{,}\) and let \(B = \set{a,b,c,d,e,f}\text{.}\) Then \(A\) is a subset of \(B\text{.}\) Why? For every \(x\text{,}\) whenever \(x\) is in \(A\) we have \(x\) is in \(B\text{.}\) Thus \(A \subseteq B\text{,}\) by definition.
Furthermore, we have \(A\) is a proper subset of \(B\text{,}\) since we additionally have that \(A\) is not equal to \(B\text{.}\) Thus \(A \subset B\text{,}\) by definition.
We apply Proof Technique Proof Technique 2.1.2.4 to show \(\emptyset \subseteq A\text{.}\) Since \(\emptyset\) has no elements, there is nothing to check. In particular, we need to check that for each \(a\) in \(\emptyset\text{,}\) the statement \(a \in A\) is true. Since there are no elements in \(\emptyset\text{,}\) there is nothing to check.
Next, we apply Proof Technique Proof Technique 2.1.2.4 to show \(A\subseteq A\text{.}\) Let \(x\) be a generic element in \(A\text{.}\) Then \(x\) is in \(A\) by construction. Thus \(A \subseteq A\text{.}\)
By definition, two sets are equal if they have the same elements. This implies that each one is a subset of the other. We can turn this into a proof technique for showing two sets are equal.
