Independent events do not necessarily have probabilities that sum to 1. All that is necessary for two events to be independent are $P(A \cap B) = P(A) P(B)$.
Perhaps you are thinking about a partition. We call a collection of events $A_1, \ldots, A_n$ a partition if $A_i \cap A_j = \emptyset$ for any $1 \leq i, j \leq n$ (that is, all the sets are disjoint) and $A_1 \cup \ldots \cup A_n = \Omega$, where $\Omega$ is the sample space. (In which case their probabilities do sum to 1.)
In my experience students confuse disjointedness and independence; that is, if they try to draw a Venn diagram representing independent events, they draw disjoint events instead. But disjoint events are not independent, because if $A$ and $B$ are disjoint and $A$ occurs, you automatically know that $B$ did not occur.
There's not much that you can say about events who's probabilities sum to one. They need not be independent, and they may not even be a partition. Probability allows for sets that are not empty but still have probability zero.