Sometimes. Proof:
Assume [tex]A\subset B[/tex].
Let [tex]a\in A\cap B[/tex], so that [tex]a\in A[/tex] and [tex]a\in B[/tex]. Then of course [tex]a\in A\cup B[/tex], so [tex]A\cap B\subseteq A\cup B[/tex].
Now let [tex]a\in A\cup B[/tex], so that [tex]a\in A[/tex] *or* [tex]a\in B[/tex]. Since [tex]A\subset B[/tex], if [tex]a\in A[/tex] then [tex]a\in B[/tex], so that [tex]a\in A\cap B[/tex]. But if [tex]a\in B[/tex], then [tex]a\not\in A[/tex], so it's not always the case that [tex]a\in A\cap B[/tex], and therefore in general [tex]A\cup B\not\subseteq A\cap B[/tex].
