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Why don't these open set axioms specify that the empty set is open?

  1. Sep 26, 2014 #1
    In all the topology text books I used in school, the open set axoims specified 4 conditions on a set S:

    (i) S is open
    (ii) empty set is open
    (iii) arbitrary union of open sets is open
    (iv) finite intersection of open sets is open


    I noticed on proofwiki, that (ii) is omitted. I was curious if anyone might be able to tell me why this is.

    https://proofwiki.org/wiki/Definition:Open_Set_Axioms

    Is it wrong, or does (ii) just follow from the other axioms in a way I don't see?
     
  2. jcsd
  3. Sep 26, 2014 #2
    If you omit (ii), it may not follow. For instance, letting ##S=\{0,1\}##, the collection of subsets ##\{\{0\}, S\}## satisfies axioms (i), (iii), and (iv).

    Of course, as long as you have a finite collection of open sets with empty intersection (e.g. two disjoint open sets), (iv) would deliver (ii).
     
  4. Sep 26, 2014 #3

    Fredrik

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    Their version of your (iii) says that the union of an arbitrary subset of ##\tau## is in ##\tau##. Since the empty set is a subset of every set, including ##\tau##, this implies that ##\bigcup\varnothing\in\tau##. Since ##\bigcup\varnothing=\varnothing##, this implies that ##\varnothing\in\tau##.

    Note that for all ##E\subseteq\tau##, ##\bigcup E## denotes the union of E. This is by definition the union of all the elements of E. So if we write ##E=\{E_i|i\in I\}##, we have ##\bigcup E=\bigcup_{i\in I}E_i##. In the case when ##E=\varnothing##, we get ##\bigcup\varnothing=\bigcup_{i\in\varnothing} E_i##. An arbitrary set x is an element of ##\bigcup_{i\in\varnothing} E_i## if and only if there's an ##i\in\varnothing## such that ##x\in E_i##. There is obviously no such i, since ##\varnothing## is the empty set. So an arbitrary x is not an element of ##\bigcup_{i\in\varnothing} E_i## and therefore not an element of ##\bigcup\varnothing##. This implies that ##\bigcup\varnothing=\varnothing##.

    Edit: This is a simpler explanation of why ##\bigcup\varnothing=\varnothing##. The left-hand side denotes the union of all elements of ##\varnothing##. Let x be an arbitrary set. The statement ##x\in\bigcup\varnothing## is equivalent to "x is an element of an element of ##\varnothing##". The latter statement is false, since ##\varnothing## doesn't have any elements. So the former statement must be false as well. Since x is arbitrary, this implies that ##\bigcup\varnothing## doesn't have any elements.

    I like your version of (iii) better.
     
    Last edited: Sep 26, 2014
  5. Sep 26, 2014 #4

    PeroK

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    The abbreviated set of axioms with (ii) omitted relies on the concept that an "arbitrary" union inclues the union of no sets. It always seemed fairly pointless to me to omit (ii) and tends to confuse the unwary. I think it's probably the author showing how clever he is rather than trying to eludicate a topic for the reader.
     
  6. Sep 26, 2014 #5
    My bad!! Listen to the other posters. :)
     
  7. Sep 26, 2014 #6
    Thanks for the responses everyone! This clears up my doubts. I must say I find this version more confusing and I'm not certain I see the benefit of writing it this way over the version with 4 axioms, which seems to be much clearer (especially for someone new).
     
  8. Sep 28, 2014 #7

    mathwonk

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    well, after initially confusing you, it did teach you something, namely that the empty union is empty. can you figure out what the empty intersection is? i.e. the intersection of an empty collection of sets, i.e. of a collection containing no subsets of X?
     
  9. Oct 7, 2014 #8
    You can also omit (i), since the null intersection the whole space.
     
  10. Oct 9, 2014 #9

    HallsofIvy

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    I remember in college my class arguing about it. The teacher finally said "yes, it is picky, but it's still true!"
     
  11. Oct 26, 2014 #10

    Stephen Tashi

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    The proof wiki takes up this issue: https://proofwiki.org/wiki/Empty_Set_is_Element_of_Topology

    That style of mathematical exposition is minimalist. Perhaps the goal is to say little enough to leave the reader confused, but enough to prove you were right if he complains. It's a good style if the objective is to give the reader a mental workout. If the effort is only to inform the reader, what deserves more explanation is the phrase "the null set". Within one "universal set", such as the "real numbers" there is a unique null set. However, .null sets are not unique in the sense that "the null set of real numbers" is not the same set as "the null set of mammals". So the symbol [itex]\emptyset [/itex] does not denote a specific set unless the context makes it clear what universal set is involved.

    The [itex] \emptyset [/itex] of [itex] \tau [/itex] is an empty subset of a collection of sets (so it is not the same set as the null set of the [itex] S [/itex] mentioned on that web page ). The ProofWiki https://proofwiki.org/wiki/Union_of_Empty_Set says that by (its) definition of the union of a collection of sets, the union of a null set collection of sets is "the null set". But is the latter null set, the null set of the collection of sets or is it the null set of the things that were elements of the sets in the collection?
     
  12. Oct 26, 2014 #11

    Fredrik

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    There's only one empty set in ZFC set theory, so for this to be an issue, you'd have to consider some other set theory, or even leave it unspecified which set theory we're going to use. If you consider a different set theory, you will have to examine its axioms. If you leave it unspecified, I doubt that the question can be answered. What would we use to answer it if not a set theory?
     
  13. Oct 26, 2014 #12

    Stephen Tashi

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    That makes me curious how ZFC deals with intersections of sets. For example if we have sets [itex] A, B [/itex] whose elements are sets of integers with [itex] A = \{ \emptyset, \{1,2\} \} [/itex] and [itex] B = \{ \{3,4, 5\}, \{10,11\} \} [/itex] then is [itex] A \cap B = \emptyset [/itex] ?
     
    Last edited: Oct 26, 2014
  14. Oct 26, 2014 #13

    Fredrik

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    Yes, because no element of A is an element of B, and no element of B is an element of A. So there's no set that's an element of both A and B.
     
  15. Nov 5, 2014 #14
    You only need to state that S is open. Or that ∅ is open. Since they are compliments of each other.
     
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