# Prove that $A\subseteq B$ if and only if $A \cap B^c = \emptyset$

Prove that $$A\subseteq B$$ if and only if $$A \cap B^c = \emptyset$$

Is this a sufficient proof for the left to right implication, i.e. proving $$A\subseteq B \implies A \cap B^c = \emptyset$$? I am focused mainly on the second paragraph where I use a contradiction.

Assume that $$A\subseteq B$$, so $$\forall x (x\in A \rightarrow x \in B)$$ is true. We show that $$A \cap B^c$$ and $$\emptyset$$ are both subsets of each other to prove equality. Note that $$\emptyset \subseteq A \cap B^c$$ is (vacously) true so now we have to show that $$A \cap B^c\subseteq \emptyset$$.

Suppose that $$x \in A \cap B^c$$. Then $$x \in A$$ and $$x \in B^c \Leftrightarrow x \notin B$$. But this contradicts with the assumption $$A\subseteq B$$ since we have $$x \in A \rightarrow x \in B$$ false. Thus $$x \notin A \cap B^c$$ for all $$x$$ and it follows that $$A \cap B^c \subseteq \emptyset$$ as $$x\in A \cap B^c \rightarrow x \in \emptyset$$ is (vacously) true. Therefore, $$A \cap B^c = \emptyset$$.

I looked here Prove that $$A \subseteq B$$ if and only if $$A \cap \overline{B}=\emptyset.$$ and the given answers there use different methods, e.g. using an expression for implication $$p \rightarrow q \iff \neg p \lor q$$ for the assumption, supposing first that $$A \cap B^c \ne \emptyset$$, etc. I am more interested in the validity of my method that I used here.

• looks good to me :) – Nosrati Oct 19 '18 at 5:08
• While your logic is correct, perhaps update the presentation. $x\in A \cap \bar B \iff x \not\in B$ is false in general, and a reader may not be paying close enough attention to the assumption that $A\subset B$. – David Peterson Oct 19 '18 at 5:10
• I think it is proved $x\in A \cap \bar B \to x\in B$ and $x \not\in B\to x\in\phi$. – Nosrati Oct 19 '18 at 5:14
• @holo Please do not use a bar on top to denote complement. This is highly non-standard. $\overset {-} B$ usually denotes the closure of $B$. – Kavi Rama Murthy Oct 19 '18 at 5:32

If A subset B, then A $$\cap$$ B$$^c$$ subset
B $$\cap$$ B$$^c$$ = empty set.
Conversely. A = (A $$\cap$$ B) $$\cup$$ (A $$\cap$$ B$$^c$$)
= A $$\cap$$ B subset B.