Error in the proof for Let $A$ and $B$ be sets, then $A^{\complement}\cup B^{\complement}\subseteq (A \cap B)^{\complement}$

Question

I have a proof that I know has a fundamental error in it, but I have no idea where the error is. Here is the proof:

Proof: Suppose that $x \in A^{\complement} \cup B^{\complement}$. This implies that $x \in A^{\complement}$ or $x \in B^{\complement}$. This disjunction will be true if $x \in A^{\complement}$. In this case, we have that $x \notin A$. But if $x \notin A$ then $x \notin A \cap B$, since $A \cap B$ is the set of all elements that are in both $A$ and $B$. And if $x \notin A \cap B$, then $x \in (A \cap B)^{\complement}$. Therefore, $A^{\complement}\cup B^{\complement}\subseteq (A \cap B)^{\complement}$ as desired.

Answer 1

Proof: Suppose that $x∈A^c∪B^c$. This implies that x∈A¯¯¯¯ or x∈B¯¯¯¯. $\color{gray}{\text{This disjunction will be true if x∈A¯¯¯¯}}$.$\color{green}{\text{Wolog we may assume, up to labeling, }x\in \overline A} $ In this case, we have that x∉A. But if x∉A then x∉A∩B, since A∩B is the set of all elements that are in both A and B. And if x∉A∩B, then x∈A∩B¯¯¯¯¯¯¯¯¯¯¯¯¯. Therefore, A¯¯¯¯∪B¯¯¯¯⊆A∩B¯¯¯¯¯¯¯¯¯¯¯¯¯ as desired.

There. Now the proof does not contain the error.

Answer 2

You have not taken into account the fact that you can have both $x\in A^{\complement}\cup B^{\complement}$ AND $x\in A$...

Since $ A^{\complement} \cup B^{\complement}$ is the addition of $A^{\complement}$ and $B^{\complement}$, so you can add back elements that were excluded from $A$ but are in $B^{\complement}$.

Answer 3

There is no fundamental error in it.

You proved legally that $x\in A^{\complement}\implies x\in(A\cap B)^{\complement}$ so that $A^{\complement}\subseteq(A\cap B)^{\complement}$.

What lacks somehow is a statement as:

"...likewise we prove that $B^{\complement}\subseteq(A\cap B)^{\complement}$ so that $A^{\complement}\cup B^{\complement}\subseteq(A\cap B)^{\complement}$".