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# Prove that the complement of an intersection is the union of the complements

Prove that .

Proof. First, let be any element in . By definition of complement, this means that and . Since we have either or which implies, coupled with the fact that is in , means or , respectively. Since is in at least one of these, is in the union . Therefore, .
For the reverse inclusion, let be an arbitrary element of . Then, either or .
If , then and ; hence, . Therefore, is in . On the other hand, if , then and ; hence, . This again implies is in . Therefore, .
Hence, .∎

### One comment

1. Anonymous says:

sorry, i cant understand Since x \notin (B \cap C) we have either x \notin B or x \notin C which
why this happens?