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Proving some relations between given sets


    \[A=\{1,2\}, \quad B=\{\{1\},\{2\}\}, \quad C=\{\{1\},\{1,2\}\}, \quad D=\{\{1\},\{2\},\{1,2\}\}.\]

Prove or disprove the following.

  1. A = B.
  2. A \subseteq B.
  3. A \subset C.
  4. A \in C.
  5. A \subset D.
  6. B \subset C.
  7. B \subset D.
  8. B \in D.
  9. A \in D.

  1. False.
    These sets are not equal since they do not contain the same elements (since 1 \neq \{ 1 \} and 2 \neq \{ 2 \}).
  2. False.
    Since A contains an element (e.g. 1) that is not contained in B, we have A \nsubseteq B. (Again, this is asking us to observe that 1 \neq \{ 1 \}.)
  3. False.
    Again, 1 \in A but 1 \notin C; hence, A \nsubseteq C.
  4. True.
    Proof. By definition of C we have \{ 1,2 \} \in C. Since A = \{1,2\}, we have A \in C.∎
  5. False.
    Since 1 \in A, but 1 \notin D, we have A \not\subset D.
  6. False.
    The element \{ 2 \} is in B, but is not in C. Hence, B \not\subset C.
  7. True.
    Proof. To show B \subset D, we must show that every element in B is also in D, and that B \neq D (so that B is a proper subset). First, the only two elements of B are the sets \{ 1 \} and \{ 2 \}. Since each of these sets is contained in D by definition of D, we have B \subseteq D. Finally, since the set \{ 1,2 \} \in D, but \{ 1,2 \} \notin B, we see that the inclusion is proper. Hence, B \subset D. ∎
  8. False.
    Again, the problem is emphasizing the difference between an element and the set that contains that element. In this case B = \{ \{1\},\{2\}\} and \{\{1\}, \{2\}\} is not an element of D (even though \{1\} and \{2\} are elements of D, the set containing the sets \{1\} and \{2\} is not).
  9. True.
    Proof. By the definition of D we have \{ 1 ,2 \} \in D. Since A = \{ 1,2 \}, we have A \in D. ∎


  1. Tesuji says:

    For\: part\: \left ( d \right ), \: \left \{ 1,2\right\left\right \} is\: an\: element\: of\: \left \{\left \{1\left\right \},\left \{2\left\right \},\left \{1,2\left\right\}\left. \right \}. So\: it\: should\: be\: true.

  2. Anonymous says:

    For part g, I think you accidentally wrote set C instead of set D starting halfway through the answer. Also, thanks for your work. Really well done.

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