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Set Proof

  1. Jun 16, 2008 #1
    PROBLEM:

    Let the (family of sets A) = {A[tex]_{\alpha}:\alpha \in \Delta}[/tex] be a family of sets and let B be a set. Prove that [tex]B \ \cup (\bigcap_{\alpha \in \Delta} A_{\alpha})\subseteq \bigcap_{\alpha \in \Delta} (B \cup A_{\alpha})[/tex]

    Don't know how to do this. Trying to get any help possible. We had a similar problem as follows:
    Let the (family of sets A) = {A[tex]_{\alpha}:\alpha \in \Delta}[/tex] be a family of sets and let B be a set. Prove that [tex]B \ \cup \bigcap_{\alpha \in \Delta} A_{\alpha} = \bigcap_{\alpha \in \Delta} (B \cup A_{\alpha})[/tex]

    To prove this we used:
    [tex]x \in \ B \ \cup \ \bigcap_{\alpha \in \Delta} A_{\alpha} \ iff \ x \in B \ or \ x \in A_{\alpha} \ for \ all \ \alpha \ iff \ x \in \bigcap_{\alpha \in \Delta} (B \cup A_{\alpha})[/tex]

    Any comments. Is this the same concept?
     
  2. jcsd
  3. Jun 16, 2008 #2
    for the first question let x be in the LHS expression and deduce that it's also in the RHS expression.

    the two questions use the same concept, but 2nd one is stronger - to prove the 2nd one you prove two inclusions, LHS is contained in RHS and RHS is contained in LHS. in the 1st one you only need to prove the the LHS is contained in the RHS.
     
  4. Jun 16, 2008 #3

    Let the (family of sets A) = {A[tex]_{\alpha}:\alpha \in \Delta}[/tex] be a family of sets and let B be a set. Prove that [tex]B \ \cup (\bigcap_{\alpha \in \Delta} A_{\alpha})\subseteq \bigcap_{\alpha \in \Delta} (B \cup A_{\alpha})[/tex]


    To prove this we used:
    [tex]Suppose \ x \in \ B \ \cup \ \bigcap_{\alpha \in \Delta} A_{\alpha} \ iff \ x \in B. \ Thus \ x \ for \ all \ A \in \alpha [/tex]

    Are you referring to setting it up as so?
     
  5. Jun 16, 2008 #4
    I would start by saying:

    Suppose x is in B union (intersection of all A's)
    then that means, either x is in B or x is in the intersection of all A's

    sorry I haven't learnt latex yet so I have to resort to just typing it all out

    and also in your setup, are you saying let x be an element of LHS iff x is an element of B? If so, then x might not be in any of the A's.
     
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