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More proofs of set relations

Let A = \{ 1 \} and B = \{ \{ 1 \}, 1 \}. Prove or disprove the following.

  1. A \subset B.
  2. A \subseteq B.
  3. A \in B.
  4. 1 \in A.
  5. 1 \subseteq A.
  6. 1 \subset A.

  1. True.
    Proof. Since 1 is the only element in A, and 1 \in B, we have that every element of A is contained in B. Hence, A \subseteq B. Further, since \{ 1 \} \in B, but is not in A, we see that B \nsubseteq A. Therefore, A \subset B. ∎
  2. True.
    Proof. Again, since A \subset B \implies A \subseteq B, we have A \subseteq B by part (a). ∎
  3. True.
    Proof. By the definition of B we have that \{ 1 \} \in B. Since A = \{ 1 \}, we have A \in B. ∎
  4. True.
    Proof. By definition A = \{ 1 \}; hence, 1 \in A.∎
  5. False.
    Since 1 is not a set, we cannot have 1 \subseteq A.
  6. False.
    Since 1 is still not a set, we cannot have 1 \subset A.

3 comments

    • Anonymous says:

      For D, we get

      *** QuickLaTeX cannot compile formula:
      {1} \in A
      
      *** Error message:
      Cannot connect to QuickLaTeX server: cURL error 60: SSL certificate problem: unable to get local issuer certificate
      Please make sure your server/PHP settings allow HTTP requests to external resources ("allow_url_fopen", etc.)
      These links might help in finding solution:
      http://wordpress.org/extend/plugins/core-control/
      http://wordpress.org/support/topic/an-unexpected-http-error-occurred-during-the-api-request-on-wordpress-3?replies=37

      if 1 is an element of A. Thus, it is true.

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