Prove that for all sets $A$ and $B$ $A\subseteq B$ implies $A\cap B=A$.

Question

In the next proof we use the following lemmas:

For $A$ and B sets, $A \subseteq B$ implies $A \cup B = B$.

For all sets $A$, $A \cap A = A$.

For all sets $B$, $B \cap \varnothing = \varnothing $.

Assume that $A⊆B$. As a result, $A \cup B = B$. Note that foregoing statement suggests two possible cases.

Case 1: $A = B$.

Since $A = B$, as a matter of fact we are trying to prove that $B\cap B=B$ which is true. Therefore, $A \cap B = A$ trivially.

Case 2: $A = \varnothing$

$A ∩ B = \varnothing ∩ B = B ∩ \varnothing = \varnothing = A.$

Then,

$A \cap B = A.$ ∎

Is this proof right?

Answer 1

This can be proved using 'modus tollens', which is:

$P\to Q \implies \lnot Q \to \lnot P$

So, we need to show that:

$$A\cap B\ne A \to A \not\subseteq B$$

If $A\cap B\ne A$, then $\exists x\in A$, such that $x\not\in B$ which means $A \not\subseteq B$, as required.

Modus tollens then states that the contrapositive, i.e. your original statement, is also true.

Answer 2

Let me add little generalization to subject, with hope to be useful. There is not only implication, but equivalence: $$A \subseteq B \stackrel{\text{(1)}} \Leftrightarrow A \cup B = B $$$$A\subseteq B \stackrel{\text{(2)}} \Leftrightarrow A \cap B = A$$

(1).

From beginning we consider first equivalence and suppose $A \subseteq B$, so $(x \in A \Rightarrow x \in B) $ is true. With this admission we have: $$x \in A \lor x \in B \Leftrightarrow (x \in A \lor x \in B) \land 1 \Leftrightarrow (x \in A \lor x \in B) \land (x \in A \Rightarrow x \in B) \Leftrightarrow \\ \Leftrightarrow (x \in A \lor x \in B) \land (x \notin A \lor x \in B) \Leftrightarrow x \in B$$ Now suppose $A \cup B = B$ and let's proof $A \subseteq B$: $$x \in A \Rightarrow x \in A \lor x \in B \Rightarrow x \in A \cup B = B$$

(2).

Again let's take $A \subseteq B$, so $(x \in A \Rightarrow x \in B) $ is true. As we know $A \cap B \subset A$ holds always, then we consider only reverse direction. In case of our admission $$x \in A \Leftrightarrow \left( x \in A \land (x \in A \Rightarrow x \in B) \right) \Rightarrow x \in B$$. And, at last,suppose we have $A \cap B = A$, then will be $$x \in A = A \cap B \subset B \Rightarrow x \in B$$