Prove that for arbitrary events A and B the following…

Prove that for arbitrary events $A$ and $B$ the following relations are equivalent: \[ \text{(i)}\ A\subseteq B; \qquad \text{(ii)}\ \overline{A}\supseteq\overline{B}; \qquad \text{(iii)}\ A+B=B; \qquad \text{(iv)}\ A\overline{B}=\varnothing. \]

Explanation of Method

"Equivalent" means that if any one of the four statements is true, all of them are true. Instead of proving all twelve implications, we prove a closed cycle (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Rightarrow$ (iv) $\Rightarrow$ (i): going around the cycle we can reach every statement from every other one. Each implication is proved by following a single outcome $\omega$ and asking which events it belongs to.

Step 1: What each statement says in words

Statement	Meaning
(i) $A\subseteq B$	whenever $A$ occurs, $B$ occurs ($A$ implies $B$)
(ii) $\overline{A}\supseteq\overline{B}$	whenever $B$ does not occur, $A$ does not occur either
(iii) $A+B=B$	"$A$ or $B$" is the same as "$B$": adding $A$ brings nothing new
(iv) $A\overline{B}=\varnothing$	$A$ can never occur without $B$

All four describe the same situation from different angles. The figure shows it:

$\Omega$ $A$ $B$ $\overline{B}$

$A$ lies inside $B$
$\overline{B}$ (grey) lies inside $\overline{A}$ (everything outside the blue circle)

$A\subseteq B$: the blue circle never reaches the grey region $\overline{B}$, so $A\overline{B}=\varnothing$ and adding $A$ to $B$ changes nothing.

Step 2: (i) $\Rightarrow$ (ii)

Assume $A\subseteq B$. Take any $\omega\in\overline{B}$, i.e. $\omega\notin B$. If $\omega$ were in $A$, then by (i) it would be in $B$ — but it is not. So $\omega\notin A$, i.e. $\omega\in\overline{A}$. Every outcome of $\overline{B}$ is in $\overline{A}$: $\overline{B}\subseteq\overline{A}$, which is (ii). ✓

Step 3: (ii) $\Rightarrow$ (iii)

Assume $\overline{B}\subseteq\overline{A}$. We show two inclusions.

$B\subseteq A+B$ — always true (a union contains each term).
$A+B\subseteq B$: take $\omega\in A+B$. If $\omega\in B$, done. If $\omega\in A$, suppose $\omega\notin B$; then $\omega\in\overline{B}\subseteq\overline{A}$, so $\omega\notin A$ — a contradiction. Hence $\omega\in B$.

So $A+B=B$, which is (iii). ✓

Step 4: (iii) $\Rightarrow$ (iv)

Assume $A+B=B$. Suppose some $\omega\in A\overline{B}$, i.e. $\omega\in A$ and $\omega\notin B$. From $\omega\in A$ we get $\omega\in A+B=B$, contradicting $\omega\notin B$. So no such $\omega$ exists: $A\overline{B}=\varnothing$, which is (iv). ✓

Step 5: (iv) $\Rightarrow$ (i)

Assume $A\overline{B}=\varnothing$. Take $\omega\in A$. If $\omega\notin B$, then $\omega\in A\overline{B}=\varnothing$, which is impossible. Hence $\omega\in B$. So $A\subseteq B$, which is (i). ✓

Step 6: Conclusion

We proved (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Rightarrow$ (iv) $\Rightarrow$ (i). Starting from any statement and following the arrows, we reach all others, so

\[ \boxed{A\subseteq B \iff \overline{A}\supseteq\overline{B} \iff A+B=B \iff A\overline{B}=\varnothing.} \]

A fifth equivalent form, proved in the same way, is $AB=A$ ("$A$ and $B$" is just "$A$").

Step 7: Example with a die

$\Omega=\{1,\dots,6\}$, $A=\{6\}$ ("a six"), $B=\{2,4,6\}$ ("an even number").

Statement	Check	True?
(i)	$\{6\}\subseteq\{2,4,6\}$	✓
(ii)	$\overline{A}=\{1,\dots,5\}\supseteq\overline{B}=\{1,3,5\}$	✓
(iii)	$A+B=\{2,4,6\}=B$	✓
(iv)	$A\overline{B}=\{6\}\cap\{1,3,5\}=\varnothing$	✓

If instead $A=\{5\}$, all four fail together: $5\notin B$, $5\in\overline{B}$ but $5\notin\overline{A}$, $A+B=\{2,4,5,6\}\neq B$, and $A\overline{B}=\{5\}\neq\varnothing$.

Result

Equivalent forms of "$A$ implies $B$"	In words
$A\subseteq B$	if $A$ occurs, $B$ occurs
$\overline{A}\supseteq\overline{B}$	if $B$ does not occur, $A$ does not occur
$A+B=B$	adding $A$ to $B$ changes nothing
$A\overline{B}=\varnothing$	$A$ never occurs without $B$