An incorrect proof by exhaustion

Question

Consider the following putative theorem.

Theorem? Suppose A, B, and C are sets and $A \subseteq B \cup C$. Then either $A \subseteq B$ or $A \subseteq C$.

What's wrong with the following proof?

Proof. Let $x$ be an arbitrary element of A. Since $A \subseteq B \cup C$, it follows that either $x \in B$ or $x \in C$.

Case 1. $x \in B$. Since x was an arbitrary element of A, it follows that $\forall x \in A(x \in B)$, which means that $A \subseteq B$.

Case 2. $x \in C$. Similarly, since x was an arbitrary element of A, we can conclude that $A \subseteq C$. Thus, either $A \subseteq B$ or $A \subseteq C$.

I have proved that the theorem is incorrect but I can't understand why the above proof is not correct.

Answer 1

When you choose cases, you lose the supposition that $x$ is arbitrary. In the first case, $x \in B$ and in the second, $x \in C$. In those respects, $x$ is not an arbitrary member of $A$.

Answer 2

It is true that for each element of $A$ you are in case 1 or case 2. But for some elements of $A$ you can be in case 1 while for others you are in case 2.

Answer 3

The error is when you switch to your cases. You start with

$$\forall x \in A:\ x \in B \text{ or } x \in C$$

and then go to cases where you act as if the above said

$$(\forall x:\ x \in B) \text{ or } (\forall x:\ x \in C).$$

See the problem?

Answer 4

If $x$ satisfies $P(x)$, then you can conclude that $\forall x\, P(x)$ is true if your $x$ was truly arbitrary, i.e. you did not make any assumptions of what your $x$ was. When considering cases, you restrict what your $x$ can be and hence the conclusion that it is true for all $x$ is false.