Proving a set is a subset of a union with itself and another set Let A and B be sets. Show that
$$\space A \subseteq (A \cup B)$$
My proof is as follows, but I don't feel confident that what I've done is correct. Any input is much appreciated.
$$x \in A \cup B \equiv x \in A \wedge x \in B \equiv x \in A$$
Therefore, A is a subset of the union of A and B.
 A: Look closely at your argument: it implies that $x\in A\cup B$ if and only if $x\in A$, which is clearly not the case in general. For instance, if $A=\{0\}$ and $B=\{1\}$, then $1\in A\cup B$, but $1\notin A$.
You’re starting in the wrong place. To show that $X\subseteq Y$, you need to show that every member of $X$ is a member of $Y$. To do this you don’t start with an arbitrary member of $Y$: some of them may not be in $X$. You start with an arbitrary member of $X$ and try to show that it must belong to $Y$. Here you want to show that $A\subseteq A\cup B$, so you should be starting with an arbitrary member of $A$. And when you do that, the rest of the argument is trivial: if $x\in A$, then it’s certainly true that $x\in A$ or $x\in B$, which is what it means to say that $x\in A\cup B$.
If you insist on doing the argument with formal logical expressions, what you want is $$x\in A\to(x\in A\lor x\in B)\leftrightarrow x\in A\cup B\;.$$
A: Hint: notice $$(x\in A\lor x\in B)\leftrightarrow x\in A\cup B\;$$ by using this logical statement $$p\Rightarrow p \lor q$$ we have  $$x\in A\Rightarrow (x\in A\lor x\in B).$$
A: You're going in the wrong direction. Don't assume $x\in A\cup B$, assume $x\in A$: you want to show that each element in $A$ is also an element of $A\cup B$ (not vice versa). Also, $x\in A\cup B\iff x\in A\textrm{ OR (not and) }x\in B$.
Let $x\in A$, and note that $x\in A\cup B\implies x\in A$ or $x\in B$. But, as $x\in A$, $x\in A\cup B$. Therefore, each element of $A$ is an element of $A\cup B$, so that $A\subseteq A\cup B$.
However, you should note that your proof is the correct proof of $A\cap B\subseteq A$ if you replace your $A\cup B$ with $A\cap B$.
A: This sort of argument is best written out in plain language, with minimal use of symbols.
Suppose $x\in A$. That is, $x$ is an element of $A$. Then certainly $x$ is either an element of $A$ or an element of $B$ (or both). Thus $x$ is an element of $A\cup B$, ie. $x\in A\cup B$.
Since we have shown that any element of $A$ is an element of $A\cup B$, $A\subseteq A\cup B$.
