Sets and subsets

  1. 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!.

  2. 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!.

  3. 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!.

  4. Determine whether each of the following statements is true or false:
    1. 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!.

    2. 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.

    3. 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!.

    4. 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!.

    5. 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!.

    6. There are no members of the set TeX Embedding failed!.

      False. The set TeX Embedding failed! contains the element TeX Embedding failed!.

    7. 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!.

    8. 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.

  5. 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!.

  6. 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.