# Statements equivalent to $A\subset B$

Each of the following statements are equivalent to $A\subset B$:

(1) $A \cap B = A$
(2) $A \cup B = B$
(3) $B^{c} \subset A^{c}$
(4) $A \cap B^{c}= \emptyset$
(5) $B \cup A^{c}= U$

I only understand (1) and (4). Why is (5) $U$ and not $B$? Can you explain/elaborate on the other ones, perhaps with a proof? Is $\subset$ used in the main statement as a subset or as a proper subset?

Thanks.

-
$\subset$ is being used here as $\subseteq$. – Brian M. Scott Jul 4 '13 at 23:04
This symbol $\subset$ denotes the proper subset while this symbol $\subseteq$ denotes just subset (which can possibly be equal to another set.) – NasuSama Jul 4 '13 at 23:05
@NasuSama: No. There is no common convention for $\subset$. That's why it's good practice to use $\subseteq$ or $\subsetneq$. (And in this question, clearly $\subset$ means $\subseteq$.) – tomasz Jul 4 '13 at 23:09

Most of them will become clearer if you draw Venn diagrams to illustrate them.

For (2), note that it’s always true that $B\subseteq A\cup B$, so what you really need to show is that $A\subseteq B$ if and only if $A\cup B\subseteq B$. Certainly if $A\subseteq B$, then $A\cup B\subseteq B$, since it’s trivially true that $B\subseteq B$. On the other hand, $A\subseteq A\cup B$, so if $A\cup B\subseteq B$, then $A\subseteq B$.

For (3), suppose that $A\subseteq B$. If $x\in B^c$, then $x\notin B$, so certainly $x\notin A$; but that means that $x\in A^c$, and we’ve shown that $B^c\subseteq A^c$. Now suppose that $B^c\subseteq A^c$. By what we just proved, we know that $(A^c)^c\subseteq(B^c)^c$. But $(A^c)^c=A$ and $(B^c)^c=B$, so $A\subseteq B$, as desired.

(5) is really very clear if you draw a Venn diagram, but you can argue as follows. Suppose that $A\subseteq B$. Let $x\in U$ be arbitrary. If $x\in B$, then certainly $x\in B\cup A^c$. If $x\notin B$, then $x\notin A$, so $x\in A^c$, and again we see that $x\in B\cup A^c$. Thus, everything in $U$ is in $B\cup A^c$, and since $B\cup A^c\subseteq U$, we must have $B\cup A^c=U$. Conversely, suppose that $B\cup A^c=U$, and let $x\in A$. Then $x\notin A^c$, but certainly $x\in U=B\cup A^c$, so it can only be that $x\in B$. This shows that $A\subseteq B$.

-

You know that $A\subset B$ is equivalent with (4).

now for realising (5), Try taking the complement on each side of (4). That is $(A\cap B^c)^c=Ø^c$

-

Intuitively, (5) means that any $x$ in the 'universe' either lies in $B$ or in the complement of $A$. In particular, if it lies in $A$ then it doesn't lie in the complement of $A$, so it must lie in $B$. That is to say, $A \subseteq B$.

Which of the other ones were you having trouble with? I don't want to spoil the fun for you.

-