Lexi Mcneil

Answered

2022-07-17

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: $(DcA)\subseteq B->A\subseteq B$, 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.

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: $(DcA)\subseteq B->A\subseteq B$, 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.

Answer & Explanation

yatangije62

Expert

2022-07-18Added 16 answers

Step 1

If A contains at least two elements, then for every element $x\in A$, we have $\{x\}\u228aA$, i.e. {x} is a proper subset of A. Therefore $\{x\}\subseteq B$ by hypothesis, i.e. $x\in B$. This means $A\subseteq B$.

Step 2

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.

If A contains at least two elements, then for every element $x\in A$, we have $\{x\}\u228aA$, i.e. {x} is a proper subset of A. Therefore $\{x\}\subseteq B$ by hypothesis, i.e. $x\in B$. This means $A\subseteq B$.

Step 2

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.

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