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Is the empty set always part of the basis of a topology?

  1. Nov 18, 2013 #1
    The topology ## T ## on a set ## X ## generated by a basis ## B ## is defined as:
    [tex] T=\{U\subset X:\forall\ x\in U\ there\ is\ a\ \beta\in B:x\in \beta \ and\ \beta\subset U \}. [/tex]
    But if ##U## is the empty set, and there has to be a ## \beta ## in ##B## that is contained in ##U##, the empty set has to be in ## B ## because only the empty set contains the empty set. Right?

    If there is something I am missing, how is the empty set part of the topology generated by a basis?
     
  2. jcsd
  3. Nov 18, 2013 #2
    No, the empty set does not need to be part of a basis (by the way, you say the basis, which is incorrect, since there could be many bases). It can be part of some bases, but it does not need to be.

    But if you take a specific basis ##\mathcal{B}## which does not contain the empty set, your question is how ##\mathcal{T}## can ever contain the empty set.

    Well, in order for ##\emptyset\in \mathcal{T}##, we must apply the definition. That is, we must show that for every ##x\in \emptyset## it must holds that there is a ##B\in \mathcal{B}## such that ##x\in B\subseteq \emptyset##. But this is always a true statement. Indeed, if the statement were false, then there would exist an ##x\in \emptyset## such that blablabla. But the empty set has no elements, so there can never exist an ##x\in \emptyset##. So a fortiori, there can not exist an ##x\in \emptyset## such that blablabla.

    These kind of statements are called vacuous statements. So any statement that says "Forall elements in the empty set holds that ..." is always true, no matter what "..." is. Any any statement that says "there exists an element in the empty set such that ..." is always false.

    So for example, every element in the empty set is a pink elephant. This is a true statement, since there are no elements in the empty set. So it's a vacuously true.
     
  4. Nov 18, 2013 #3
    What you are saying is that any statement that says "For all elements in the empty set some property X holds" has to be true because the negation of it, "There exists an element in the empty set such that X does not hold", is always false, since there is no element in the empty set?

    Meaning something is true, even if not obviously, because it's negation is obviously false (because saying that a statement is correct is the same as saying that its negation is false, right?). And something that is true by that logic is said to be vacuously true?

    Specifically in this case, you turned the problem into trying to disprove that there is a ##x\in \emptyset## that is not in some ##B\in \mathcal{B}## such that ##B\subseteq \emptyset##. Which is automatically false because there is no elements in the empty set.
     
  5. Nov 18, 2013 #4
    Yes, that's what I mean.
     
  6. Nov 18, 2013 #5
    :D Thanks, just checking that I understood!
     
  7. Nov 18, 2013 #6
    Wait, ins't that paradoxical?

    I mean, if the proposition "for all elements in A some property X holds" is true, then shouldn't the veracity of the expression "there is an element in A for which property X holds" be a consequence of it? But the second is false in the case that A is the empty set, because there are no elements in the empty set. Does the first logical expression not imply the second? Or is there something else I am missing?
     
  8. Nov 18, 2013 #7
    Why would "All elements in A have property X" imply that "There is an element of A with property X"?
     
  9. Nov 18, 2013 #8
    Okay, I now see that it doesn't, sorry.. This is sorta confusing when you're a noob :)

    So conversely if "there is an element in A for which property X holds" is false and "all elements in A have property X" is true, then A has to be empty, right?
     
  10. Nov 20, 2013 #9
    One more question, if two elements in the basis collection do not intersect, does that mean that the empty set is in the basis collection as well?

    I've seen two definitions for the collection of basis elements, both have two conditions. The first condition coincided, but the second was stated differently, as follows:

    If there is a ##x\in (B_1\cap B_2)## for some ##B_1## and ##B_2## in the basis, than there is a ##B_3## in the basis such that ##x\in B_3\in (B_1\cap B_2)##.

    For all ##x\in (B_1\cap B_2)## for some ##B_1## and ##B_2## in the basis, there is a ##B_3## in the basis such that ##x\in B_3\in (B_1\cap B_2)##.

    Wouldn't the empty set satisfy the second version vacuously? But not satisfy the first at all? Which one is the correct definition.

    I guess a clue is that the first comes from Munkres text book on point set topology, the second wikipedia :D
     
  11. Nov 20, 2013 #10
    Both versions mean exactly the same thing. If ##B_1\cap B_2# is empty, then both conditions are satisfied vacuously.
     
  12. Nov 20, 2013 #11
    The statement "if there is A then there is B" means the same as "for all A there is B"?

    And also, is it right that if two elements in the basis collection do not intersect, the empty set is in the basis collection as well?
     
  13. Nov 20, 2013 #12

    Office_Shredder

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    Let's suppose that [itex] B_1, B_2[/itex] are in the basis and do not intersect, and try each of your conditions:
    1.) If there is an x in [itex] B_1\cap B_2[/itex], then we know about the existence of a [itex] B_3[/itex]. But there is no such x, so we don't know that a [itex] B_3 \subset B_1 \cap B_2[/itex] exists.

    2.) For every [itex] x\in B_1 \cap B_2 [/itex], we know there exists a [itex] B_3[/itex]. There are no x's, so we learn about the existence of zero new sets. As said before, the statement "for all x in the empty set, ______" holds regardless of what ___ is. We don't need to change the basis to make _____ hold (in this case the existence of a B3), because we knew the statement was true as soon as we talked about "for all x in the empty set".
     
  14. Nov 20, 2013 #13
    I asked my professor this same question, but for a different reason. I asked because we learned that any member of the topology can be written as a union of basis elements. Then, how do we get the empty set through unions without the empty set? And the answer to that is that the empty union is the empty set.

    Just an added reason why we don't need empty sets in our bases.
     
  15. Nov 20, 2013 #14
    And what would the empty intersection be? :devil:
     
  16. Nov 20, 2013 #15
    Well that would just be the point set, but why?
     
  17. Nov 20, 2013 #16
    It's actually undefined. The only thing that would make sense is the set of all sets, but that doesn't exist.

    Why? Because it's fun to reason about such concepts :tongue:
     
  18. Nov 20, 2013 #17
    If we are talking about a topology on a set X, then we've taken X to be our universal set. Why is it undefined? X would be the identity element under intersections in all cases, right?
     
  19. Nov 20, 2013 #18
    Right, but the notion of universal set has no real meaning in set theory. So when taking ##\bigcap \emptyset##, then we usually work with the ZFC-axioms. Under that axiom system, we haven't defined a universal set. So there is no way to specify you are taking an empty intersection that is "in X".

    Note however, that you can give meaning to universal sets very easily. But that is rarely, if ever, done.

    I think that empty intersection never really come up anyways... The only reason to work with universal sets is that complements become easier...
     
  20. Nov 20, 2013 #19
    In the Aristotle theory of syllogisms it is a valid categorical syllogism that "all S are P" implies his subaltern "some S are P". But that theory don't consider the possibility that "there is no S at all", ie, this theory don't consider the idea of an empty set. The arguments showed in this topic about the empty set are valid and may sound strange at first sight.
    Even today in logic there are some counter intuitive ideas, like the material implication as an example. In formal logic when studying models the student meet the "structures" that have to be nonempty sets in order to some sentences make sense.
     
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