- Prove that for any set
*TeX Embedding failed!*,*TeX Embedding failed!*Assume

*TeX Embedding failed!*then*TeX Embedding failed!*, but this is impossible because*TeX Embedding failed!*is empty, hence our assumption is false and it is true that*TeX Embedding failed!*. - Prove that
*TeX Embedding failed!*Assume

*TeX Embedding failed!*then*TeX Embedding failed!*and*TeX Embedding failed!*. Now*TeX Embedding failed!*but*TeX Embedding failed!*because*TeX Embedding failed!*contains nothing, therefore*TeX Embedding failed!*so the assumption is false and it is true that*TeX Embedding failed!*. - Prove that for any set
*TeX Embedding failed!*,*TeX Embedding failed!*.Clearly

*TeX Embedding failed!*so that*TeX Embedding failed!*.*TeX Embedding failed!*is called an improper subset of*TeX Embedding failed!*. - Determine whether each of the following statements is true or false:

- For each set
*TeX Embedding failed!*,*TeX Embedding failed!*.True. By definition

*TeX Embedding failed!*is the power set of*TeX Embedding failed!*and contains all subsets of*TeX Embedding failed!*. Since*TeX Embedding failed!*it follows that*TeX Embedding failed!*. - For each non-empty set
*TeX Embedding failed!*,*TeX Embedding failed!*.%False. For

*TeX Embedding failed!*to be true it must hold that if*TeX Embedding failed!*then*TeX Embedding failed!*. But if*TeX Embedding failed!*is any non-empty set it contains elements and an element of*TeX Embedding failed!*is not a subset of*TeX Embedding failed!*, it is therefore not in*TeX Embedding failed!*and hence*TeX Embedding failed!*.False. Assume

*TeX Embedding failed!*, then*TeX Embedding failed!*. Consider any such*TeX Embedding failed!*,*TeX Embedding failed!*but*TeX Embedding failed!*. Therefore the assumption, namely that*TeX Embedding failed!*, is false. - For each set
*TeX Embedding failed!*,*TeX Embedding failed!*.True.

*TeX Embedding failed!*contains only the element*TeX Embedding failed!*and*TeX Embedding failed!*so*TeX Embedding failed!*. Everything in*TeX Embedding failed!*is thus in*TeX Embedding failed!*so it follows from the definition of subset that*TeX Embedding failed!*. - For each set
*TeX Embedding failed!*,*TeX Embedding failed!*.True. It was proven above that for any set

*TeX Embedding failed!*,*TeX Embedding failed!*.*TeX Embedding failed!*contains all subsets of*TeX Embedding failed!*by definition,*TeX Embedding failed!*, therefore*TeX Embedding failed!*.%Since

*TeX Embedding failed!*contains all subsets of*TeX Embedding failed!*it clearly contains*TeX Embedding failed!*. - For each set
*TeX Embedding failed!*,*TeX Embedding failed!*.True.

*TeX Embedding failed!*is a subset of any set*TeX Embedding failed!*therefore*TeX Embedding failed!*irrespective of*TeX Embedding failed!*. - There are no members of the set
*TeX Embedding failed!*.False. The set

*TeX Embedding failed!*contains the element*TeX Embedding failed!*. - Let
*TeX Embedding failed!*and*TeX Embedding failed!*be sets. If*TeX Embedding failed!*, then*TeX Embedding failed!*.%True. Assume that it is not true that if

*TeX Embedding failed!*, then*TeX Embedding failed!*. It follows that there exists a subset*TeX Embedding failed!*of*TeX Embedding failed!*which is not a subset of*TeX Embedding failed!*, but since any element in*TeX Embedding failed!*is also in*TeX Embedding failed!*, this is impossible. By the same construction steps that we apply to the elements of*TeX Embedding failed!*to construct*TeX Embedding failed!*we can construct*TeX Embedding failed!*from the elements of*TeX Embedding failed!*as they occur in*TeX Embedding failed!*. Therefore the assumption is false and it is true that if*TeX Embedding failed!*, then*TeX Embedding failed!*.True. Let

*TeX Embedding failed!*and assume*TeX Embedding failed!*. By the assumption,*TeX Embedding failed!*. Consider this*TeX Embedding failed!*, by definition of*TeX Embedding failed!*,*TeX Embedding failed!*and therefore*TeX Embedding failed!*but*TeX Embedding failed!*which means*TeX Embedding failed!*, therefore*TeX Embedding failed!*which by definition means that*TeX Embedding failed!*.Now if

*TeX Embedding failed!*then by definition of*TeX Embedding failed!*,*TeX Embedding failed!*but this is a contradiction since it was claimed that*TeX Embedding failed!*. Therefore the assumption, namely that*TeX Embedding failed!*given*TeX Embedding failed!*, is false, and it is therefore true that if*TeX Embedding failed!*, then*TeX Embedding failed!*. - There are two distinct objects that belong to the set
*TeX Embedding failed!*.True since

*TeX Embedding failed!*and*TeX Embedding failed!*are distinct objects and they belong to the set.

- For each set
- Let
*TeX Embedding failed!*,*TeX Embedding failed!*,*TeX Embedding failed!*be sets. Prove that if*TeX Embedding failed!*and*TeX Embedding failed!*then*TeX Embedding failed!*.If

*TeX Embedding failed!*, then*TeX Embedding failed!*, and if*TeX Embedding failed!*, then*TeX Embedding failed!*. It follows from the former, given the latter, that*TeX Embedding failed!*, and hence*TeX Embedding failed!*.Alternative proof: If

*TeX Embedding failed!*then*TeX Embedding failed!*, and if*TeX Embedding failed!*then*TeX Embedding failed!*. Thus*TeX Embedding failed!*and hence*TeX Embedding failed!*thus*TeX Embedding failed!*.Alternative proof (by contradiction): Assume that

*TeX Embedding failed!*and*TeX Embedding failed!*but*TeX Embedding failed!*, this means that*TeX Embedding failed!*. Taking this*TeX Embedding failed!*since*TeX Embedding failed!*,*TeX Embedding failed!*, and since*TeX Embedding failed!*,*TeX Embedding failed!*which is a contradiction. Therefore the assumption is false, namely that*TeX Embedding failed!*, and hence it is true that if*TeX Embedding failed!*and*TeX Embedding failed!*, then*TeX Embedding failed!*. - Let
*TeX Embedding failed!*be sets. Prove that if*TeX Embedding failed!*and*TeX Embedding failed!*, then*TeX Embedding failed!*.Proof by induction on

*TeX Embedding failed!*.The base case,

*TeX Embedding failed!*is trivially true since*TeX Embedding failed!*. %For the base case*TeX Embedding failed!*,*TeX Embedding failed!*and*TeX Embedding failed!*, therefore by definition of set equality*TeX Embedding failed!*.Now it is required to show that for any

*TeX Embedding failed!*, if the claim holds for*TeX Embedding failed!*then it holds for*TeX Embedding failed!*. For*TeX Embedding failed!*we have*TeX Embedding failed!*and

*TeX Embedding failed!*. Now*TeX Embedding failed!*and*TeX Embedding failed!*which means that*TeX Embedding failed!*, therefore*TeX Embedding failed!*and*TeX Embedding failed!*both hold. Given that the claim holds for*TeX Embedding failed!*it follows that*TeX Embedding failed!*and therefore because

*TeX Embedding failed!*that*TeX Embedding failed!*because*TeX Embedding failed!*. Given that it is also true that*TeX Embedding failed!*, this means that*TeX Embedding failed!*and since*TeX Embedding failed!*it follows that*TeX Embedding failed!*and the induction step holds.