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I Why is the empty set a proper subset of every set?

  1. Sep 16, 2016 #1
    I know what a proper subset is, but I never understood why every set has the empty set as its subset?

    I mean, is the reasoning something primitive like this: if I have x objects, the number of unordered sets of elements I can make are 2^x, including the case where I throw out x objects and dont make anything of them?
     
  2. jcsd
  3. Sep 16, 2016 #2

    TeethWhitener

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    S is a subset of A iff all elements of S are elements of A. Since the empty set has no elements, this condition is trivially satisfied: the empty set is a subset of all sets. S is a proper subset of A iff S is a subset of A and S is not equal to A. The empty set is therefore a proper subset of any non-empty set.
     
  4. Sep 16, 2016 #3

    mfb

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    The "non-empty" is important here. The empty set is not a proper subset of the empty set. It is still a subset.
     
  5. Sep 16, 2016 #4
    I dont understand why
     
  6. Sep 16, 2016 #5
    A set A is a proper subset of another B set if and only if all elements of A are contained within B, and B has at least one element that is not contained in A.

    The ampty set is a proper subset of all sets except itself.

    But why?
     
  7. Sep 16, 2016 #6

    TeethWhitener

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    Are all elements of ##\emptyset## elements of ##A##?
     
  8. Sep 16, 2016 #7
    There is no intersection, it is a disjoing, but A contains as the empty subset as a subset....so there is an intersection? Now i am confused
     
  9. Sep 16, 2016 #8

    TeethWhitener

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    Let's think about it a different way: If, for all ##x \in S##, it's also true that ##x## is an element of ##A##, then ##S \subseteq A##. So now all you have to do is list out all of the ##x \in \emptyset##, and check to make sure that they're all in ##A##.

    EDIT: this may require a little knowledge of first-order logic. In symbolic terms, the condition for ##S \subseteq A## is
    $$(S \subseteq A) \leftrightarrow \forall x (x \in S \rightarrow x \in A)$$
    The rule for material implication is that ##p \rightarrow q## is true when ##p## is false, regardless of what ##q## is. So ##x \in \emptyset## is false for all ##x##, which means the condition $$\forall x (x \in S \rightarrow x \in A)$$
    evaluates to
    $$\forall x (F \rightarrow x \in A)$$
    $$\forall x (T)$$
    $$T$$
    where ##F## is "false" and ##T## is "true."
     
    Last edited: Sep 16, 2016
  10. Sep 16, 2016 #9

    mfb

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    Let's take a different approach: "Every square number between 5 and 7 is odd". Is this statement correct?

    "Every element in ∅ is also element of set A". Is this statement correct?
    Same logic.
     
  11. Sep 16, 2016 #10
    All I know about logic is this:

    SMl5t4c.png

    mISTAKE/typo..UNION Is the OR operator, but yes, this is all I know
     
  12. Sep 16, 2016 #11

    TeethWhitener

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    We can try a "proof by contradiction" as well (even though this is a definition, not a theorem). Assume that the empty set isn't a subset of ##A##. Then there exists some element of the empty set that is not an element of ##A##. But this is impossible, because the empty set has no elements. So the empty set has to be a subset of ##A##.
     
  13. Sep 16, 2016 #12

    micromass

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    Let ##A## be a non-empty set. Can you name me an element of ##\emptyset## that is not in ##A##?
     
  14. Sep 16, 2016 #13
    No, and that is why it must be a subset of A.:oldbiggrin:
     
  15. Sep 16, 2016 #14

    mfb

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    Correct.
    This is not logic. This is about unions and intersections of sets, which do not matter here.
     
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