# If $A_1\cap…\cap A_n \neq \emptyset$, does $(A_1\cap…\cap A_n)^{c} =A_1^{c} \cup … \cup A_n^{c} = \emptyset$?

If I have some collection of sets such that $A_1\cap...\cap A_n \neq \emptyset$, then what happens if I apply the complement (denoted by superscript c) to both sides? i.e.,

$(A_1\cap...\cap A_n)^{c} =A_1^{c} \cup ... \cup A_n^{c} = \emptyset?$

Thank you!

## 3 Answers

Your claim is not true. The complement of a nonempty set is not necessarily the empty set.

An example:

Let $X = \{a,b,c\}$. Let $A_1 = \{a,b\}$ and $A_2 = \{b, c\}$.

$A_1 \cap A_2 = \{b\} \neq \emptyset$.

However $(A_1 \cap A_2)^c = A_1^c \cup A_1^c = \{a,c\} \neq \emptyset$.

No. If $\bigcap_{i=1}^nA_i\neq\varnothing$, then all you can conclude is that its complement is not the whole space. That is, if there exists some $x\in \bigcap_{i=1}^nA_i$, then for this $x$ it is also true that $x\notin \bigcup_{i=1}^n A_i^c$.

It depends, consider $A_i = X$ the whole set, or $A_i = \{i\}$ as subset of $\mathbb{N}$