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: ( D c A ) ⊆ B − &gt; 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.

Question

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&#039;m not sure how you would write this using symbols. Something like:   (  D  c  A  )  ⊆  B  −  &amp;gt;  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.

yatangije62 · Accepted Answer

Step 1If A contains at least two elements, then for every element   x  ∈  A, we have   {  x  }  ⊊  A, i.e. {x} is a proper subset of A. Therefore   {  x  }  ⊆  B by hypothesis, i.e.   x  ∈  B. This means   A  ⊆  B.Step 2If 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.