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$$A \cap B \subseteq A$$

My first step would be to write it as $(x \in A \land x \in B) \subseteq A$. Then I know by the following implication that is always true $P \land Q \implies P$. But I am not sure how to write it down mathematically correct.

$$A \subseteq A \cup B $$

I would write it as $A \subseteq (x \in A \lor x \in B)$ . Then I know by following implication that is always true $P \implies P \lor Q$. But again I do not know how to write in down mathematically correct.

May you could help a beginner?

Kahnu
  • 51

3 Answers3

2

$x \in A$ is a logical expression and so is $\left((x \in A) \land (x \in B)\right)$. The operator $\subseteq$ operates on sets, the result is a logical value. The definition is

  • $U \subseteq V$ if and only if $(x \in U) \implies (x \in V)$

The other definitions you use ar

  • $ x \in U \cap V$ if and only if $ (x \in U) \land (x \in V)$

  • $ x \in U \cup V$ if and only if $ (x \in U) \lor (x \in V)$

So the expression $((x \in A) \land (x \in B)) ⊆ A$ does not make sense because you put a logical operator on the left side of $\subseteq$.

$P$ and $Q$ are logical variables in

$$P \land Q \implies P$$

If you substitute $P$ by $(x \in A)$ and substitute $Q$ by $(x \in B)$ you get

$$(x \in A) \land (x \in B) \implies (x \in A)$$

But the left side of the implication is the definition of $x \in (A \cap B)$, so you get $$x \in (A \cap B) \implies (x \in A)$$ which again is the definition of $\subseteq$.

Your second statement can be proved in a similar way.

miracle173
  • 11,049
1

You are correct to start with let $x\in A\cap B$ so $x\in A\text{ and } B$. Since x is in A we know that $A\cap B\subseteq A$.

Now let $x\in A$. By definition, x is in the union if it is in A or B so $A$ is a subset of the union.

dustin
  • 8,241
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To show this, show that any element $x\in A\cap B$ (read as "x in A intersection B") must be in A, and similarly for the union, show that any $x\in A$ must be in $A\cup B$.

Suzu Hirose
  • 11,660