I am very lost on this question for my discrete math class:
Let A and B be sets. Suppose A contains at least 2 elements. Prove that if every proper subset of A is a subset of B, then A is a subset of B. (Hint: what does it mean for a subset of A with size one to be a subset of B?)
What is an example and explanation to demonstrate why the above proof can be false if A contains only 1 element.
When I attempted this question on my own, I couldnt even get started. I follow the one answer that was already provided but I'm not sure how you would write this using symbols.
Something like: (DcA)⊆B->A⊆B, xED so could be XEA but D does not equal A. If xEA, Then xEB. But this reasoning must be flawed as I am not taking into account the hint.
Best Answer
If $A$ contains at least two elements, then for every element $x \in A$, we have $\{x\} \subsetneq A$, i.e. $\{x\}$ is a proper subset of $A$. Therefore $\{x\} \subseteq B$ by hypothesis, i.e. $x \in B$. This means $A \subseteq B$.
If you $A$ contains only one element, then the only proper subset of $A$ is empty, so you simply have two sets $A$ and $B$ with no conditions whatsoever on them, so there is no reason for $A$ to be a subset of $B$.
Hope that helps,