Proving if $A \subseteq B$ and $A \nsubseteq C$ then $B \nsubseteq C$

Question

This is one of the problem I have been solving in Velleman's How to prove book:

Prove that if $A \subseteq B$ and $A \nsubseteq C$ then $B \nsubseteq C$

This is my solution:

Suppose $A \subseteq B \cap A \nsubseteq C$. Let x be an arbitrary element in A. From $x \in A$ and $A \subseteq B$, it follows that $x \in B$. From $x \in A$ and $A \nsubseteq C$, it follows that $x \notin C$. Therefore, $x \in B$ and $x \notin C$. Since x is arbitrary, $B\nsubseteq C$.

This is the solution given in the book:

Suppose that $A \subseteq B$ and $A \nsubseteq C$. Since $A \nsubseteq C$, we can choose some $a \in A$ such that $a \notin C$. Since $a\in A$ and $A \subseteq B$, $a \in B$. Since $a \in B$ and $a \notin C$, $B \nsubseteq C$.

The book uses existential quantification to prove it but I assume x as arbitrary and prove that. Is there some flaw in my way of thinking ?

Answer 1

The flaw in your solution is "from $x\in A$ and $A\not\subseteq C$, it follows that $x\not\in C$." This may not be true for every $x\in A$, but the hypothesis says that this is true for at least one. This is why the solution that you are given uses existential quantification.

For a concrete example, consider $A=\{0,1\}, B=\{0,1,2\}, C=\{0\}$. It is not true that every $x$ in $A$ does not belong to $C$, but this is true for at least one.

Answer 2

A proof of the problem may be obtained if you assume $A\subseteq B \land B\subseteq C\land A\not\subseteq C$. Because, $$(A\subseteq B \land B\subseteq C)\land A\not\subseteq C\implies A\subseteq C \land A\not\subseteq C$$ Which surely is a contradiction.

Answer 3

If possible suppose $B\subseteq C$.AS $A\subseteq B$,that would imply $A\subseteq C$,which is a contradiction.