Why does $ x \in (A \cap B)^c \implies x \in A^c \cup B^c $

elementary-set-theory

The following is true:

$$ \boxed{ x \in (A \cap B)^c \implies x \in A^c \cup B^c }$$

Notation – here $S^c$ is the complement of the set $S$.


Question: Why is this true?

I know it is true by constructing a Venn diagram to show that it is.

However I would like to see a step-by-step argument based on the fundamental definitions of sets and the union/intersection/complement operators.

This question also asks the same question but the comments and answers fail to provide a solution to the specific question of why.


Further research:

Understanding Analysis, Second Edition, by the well respected Stephen Abbott, has an exercise and an official solution by the author. To quote the step I am not understanding:

If $x \in (A \cap B)^c$ then $x \notin (A \cap B)$. But this implies
$x \notin A$ or $x \notin B$.

Why does this imply $x \notin A$ or $x \notin B$ ? This is not obvious to me.

Again, to clarify, I can draw Venn diagrams to confirm the truth but I would like to see logical steps based on fundamental definitions or axions if possible.

Attempting to think about this in plain English with a real-world example to see if it helps .. doesn't seem to help:

If it is not summer and hot, then it is not summer or not hot.

Best Answer

If it were true that $x \in A \cap B$, then $x \in A$ and $x\in B$. That is, $$ x \in A \cap B \iff (x \in A \land x \in B) $$ Well, $x \not \in A \cap B$, so we consider the negation: $$ x \not \in A \cap B \iff \neg (x \in A \cap B) \iff \neg (x \in A \land x \in B) $$ Then we ask: what is the negation of a logical "and" statement, say $\neg (P \land Q)$? It is, as be shown in a way of your preference (e.g. truth tables) that $$ \neg (P \land Q) \equiv (\neg P) \lor (\neg Q) $$ Hence $$ x \not \in A \cap B \iff \underbrace{x \not \in A}_{\neg (x \in A)} \lor \underbrace{x \not \in B}_{\neg (x \in B)} $$

If the shape is not blue and circular, then either it is not blue, or it is not circular. (Or both. The logical "or" is inclusive.)

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