Prove that $B ⊆ A^c ⇔ A ∩ B = ∅$

Question

The question I am trying to answer is:

Prove that $B ⊆ A^c ⇔ A ∩ B = ∅$

This is what I have:

First assume $B ⊆ A^c$. If $x∈A∩B$, then $x∈A$ and $x∈B$. However, by assumption $B ⊆ A^c=U-A$ for a universe $U$. So, $B ⊆ U-A$ which means $x∈B ⊆ U-A$ and $x∈B$ but $x∈/A$. Thus, $∅=A ∩ B$. Now assume A ∩ B = ∅.

This is where I got stuck. I was able to prove $B ⊆ A^c => A ∩ B = ∅$ though how do you prove $A ∩ B = ∅ => B ⊆ A^c$ ? I'm also not sure if what I did write is correct. Can someone check if I made any errors and explain how I would complete this proof?

Answer 1

Lemma 1

The following equivalence holds: $X\subseteq Y \Longleftrightarrow X\cap Y = X$.

Proof

Let us prove the implication $(\Rightarrow)$ first.

The sets $X\cap Y$ and $X$ are equal iff $X\cap Y\subseteq X$ and $X\subseteq X\cap Y$.

If $a\in X\cap Y$, then, according to the definition of intersection, one has that $x\in X$.

Conversely, if $a\in X$ and $X\subseteq Y$, then $a\in Y$. That is to say, $a\in X\cap Y$, and we are done.

Let us now prove the implication $(\Leftarrow)$.

If $X\cap Y = X$ and $a\in X$, then $a\in X\cap Y$, that is to say, $a\in X$ and $a\in Y$.

Thus we have proven that $X\subseteq Y$ and we are done.

Lemma 2

If $X\subseteq W$ and $Y\subseteq W$, then $X\cup Y \subseteq W$.

Proof

Indeed, if $a\in X\cup Y$, then $a\in X$ or $a\in Y$. If $a\in X$, then $a\in W$.

Similarly, if $a\in Y$, then $a\in W$. In both cases, $a\in W$.

Whence we conclude that $X\cup Y\subseteq W$.

Lemma 3

The following equivalence holds: $X\subseteq Y \Longleftrightarrow X\cup Y = Y$

Proof

Let us prove the implication $(\Rightarrow)$ first.

The sets $X\cup Y$ and $Y$ are equal iff $X\cup Y\subseteq Y$ and $Y\subseteq X\cup Y$.

The inclusion $Y\subseteq X\cup Y$ is trivial, since $a\in Y$ implies that $a\in X$ or $a\in Y$.

That is to say, $a\in X\cup Y$.

Conversely, if $X\subseteq Y$, then $X\cup Y = Y$, because $Y\subseteq Y$, and we are done.

Let us prove the implication $(\Leftarrow)$ now.

If $a\in X\cup Y = Y$, then $a\in X$ or $a\in Y$. Since $X\cup Y\subseteq Y$, then $a\in Y$.

Thus we conclude that $X\subseteq Y$, just as desired.

Proposition

It is true that $B\subseteq A^{c} \Longleftrightarrow A\cap B = \varnothing$

Proof

Let us prove the implication $(\Rightarrow)$ first.

Based on the previous result, we have that \begin{align*} B\subseteq A^{c} \Longrightarrow B\cap A^{c} = B & \Longrightarrow A\cap B = A\cap(B\cap A^{c}) = \varnothing \end{align*}

Conversely, let us prove the implication $(\Leftarrow)$.

If $A\cap B = \varnothing$, then we have that \begin{align*} A\cap B = \varnothing \Longrightarrow (A\cap B)\cup A^{c} = B\cup A^{c} = \varnothing\cup A^{c} = A^{c} \Longrightarrow B\subseteq A^{c} \end{align*}

Hopefully this helps.

Answer 2

Suppose for contradiction that $B$ is not subset of $A^{c}$. Then there exist $x \in B$, s.t $x \notin A^{c}$. Since $x \notin A^{c}$, we have $x \in A$. Then $x \in B$ and $x \in A$, that is, $x \in A \cap B$. It is a contradiction, because $A \cap B = \emptyset.$

Answer 3

Let $B \subseteq A^c$. As $A \cap A^c = \emptyset$, we immediately have $A \cap B = \emptyset$. On the other hand, if $A \cap B = \emptyset$, then if $x \in B$ we must have $x \not\in A$. So in some larger universe $U$ this says that $x \in B \Rightarrow x \in U\setminus A = A^c$. Thus, $B \subset A^c$.

Answer 4

For the converse part follow this Let $ x\in B$ and by given condition we can say $ x\notin A$ Thus $x \in A^c$ this implies that $ B\subset A^c$

Answer 5

$(\Rightarrow)$

Let $B \subseteq A^{c}$.

Assume $A \cap B \neq \emptyset$. [for contradiction]

$A \cap B \neq \emptyset \implies$ there exists an $x$ in both $A$ and $B$.

$x \in A \implies x \notin A^{c}$

$x \in B \implies x \in A^{c}$ [since $B \subseteq A^{c}$]

This is a contradiction since $x$ cannot be both in $A^{c}$ and not in it.

So $A \cap B = \emptyset$.

$(\Leftarrow)$

Let $A \cap B = \emptyset$.

$A \cap B = \emptyset \implies $ for all $x \in B$, $x \notin A$.

$x \notin A \implies x \in A^{c}$

$x \in A^{c}$ for all $x\in B \implies B \subseteq A^{c}$