If $A\subseteq B$, can $A$ and $B$ be independent?

I know that in probability theory, to say that two events are independent means that the occurrence of one does not affect the probability of the other.

But if $A$ is a subset of $B$, then $A$ and $B$ are not independent right? Also I do not understand the second part at all. Can somebody help me out?

Independence means $P(AB)=P(A)P(B)$.

So, if $A\subseteq B$, then $AB=A$ and so $P(AB)=P(A)$. So, $A$ and $B$ are independent precisely when $P(A)=P(AB)=P(A)P(B)$.

If $P(A)=0$ or $P(B)=1$, then the equality holds, hence $A$ and $B$ are independent (so the answer to the first question is "yes, they can be independent").

However, if $P(A)\ne 0$ then the equality $P(A)=P(A)P(B)$ would yield $P(B)=1$. Hence, the answer to the second question is "no, that is impossible".

Hint: By definition, events $A$ and $B$ are independent when $$P(A\cap B)=P(A)P(B).$$ Note that any probability is always greater than or equal 0, and less than or equal to 1. Also note that because $A\cap B\subseteq A$, we have $P(A\cap B)\leq P(A)$, and similarly, because $A\cap B\subseteq B$, we have $P(A\cap B)\leq P(B)$.

Do you see where the relevance of allowing (or not allowing) $P(A)$ and $P(B)$ to be 0 or 1 comes in now? Try letting $A$ be $\varnothing$, for example, or let $B=\Omega$ (the whole event space).

To add another insight, note also that when events $A$ and $B$ are independent, we must have $$P(B|A)=P(B).$$ Also,$$P(B|A)=\frac{P(A\cap B)}{P(A)}.$$ However, since $A\subseteq B$, then $A\cap B= A$ and consequently $P(A\cap B)=P(A)$. Thus for the two events to be independent, we must have $P(B|A)=P(B)=1$.