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Why is the empty set open?

  1. May 18, 2008 #1
    The definition of an open set S is that there exists, for any point x element of S, an open ball with center x, all of whose points belong to S.

    But if S is the empty set, then it contains no point. How can we say that the above is valid if there is no point around which to build an open ball?

    I can understand why the empty set is closed. Since any closed set contains all its accumulation points and since the empty set has no such accumulation point, the set of all accumulation points is empty and is therefore contained (an equal!) to the empty set.

    But in the case of openness, we are faced with a P => Q where P is not true. In such case, we cannot say anything about Q, so to me, the question of whether the empty set is open or not has no answer...

    I'd appreciate your mathematical wisdom on this one...:smile:

    PS. After more thought, I have come to a possible "explanation". Since any statement is either true or false and since the statement that the empty set is open clearly is not false (as in not falsifiable), then it must be true. (Of course, I could also say that it is not true, so it must be false...)
    Last edited: May 18, 2008
  2. jcsd
  3. May 18, 2008 #2


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    From wikipedia: http://en.wikipedia.org/wiki/Empty_set

    Considered as a subset of the real number line (or more generally any topological space), the empty set is both closed and open. All its boundary points (of which there are none) are in the empty set, and the set is therefore closed; while for every one of its points (of which there are again none), there is an open neighbourhood in the empty set, and the set is therefore open. Moreover, the empty set is a compact set by the fact that every finite set is compact.
  4. May 19, 2008 #3


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    I'm confused.

    If the empty set is open, then its complement must be closed.

    But the complement of the empty set is the whole set, which needn't be closed. :confused:
  5. May 19, 2008 #4


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    An open set is one that, for all x in the set, blah blah blah. [tex]\forall x\in\emptyset\varphi(x)[/tex] is true for all predicates [tex]\varphi[/tex] since there aren't any members to falsify it.

    1. If a set is open, its complement is closed.
    2. The empty set is open.
    3. The complement of the empty set is closed.
    4. The complement of the empty set need not be closed.

    Steps 1 and 2 are true; step 3 follows. Step 4 is false.
    Last edited: May 19, 2008
  6. May 19, 2008 #5
    But it wouldn't be a contradiction to say: "for every one of its points (of which there are again none), there is no open neighbourhood in the empty set, and the set is therefore closed."

    There is no point around which to build a n-ball so any statement drawn from the use of such n-balls is bound to be inconclusive. At least, that's how I see it...

  7. May 19, 2008 #6


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    Why is this a problem? The definition is "for all x in the set, stuff", not "for all x in the set, stuff, plus there is some x in the set". Every member of the empty set is a pony. Every member of the empty set is an inaccessible cardinal. Every member of the empty set is your father.
  8. May 19, 2008 #7


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    The negation of
    [tex]\forall x\in S,\varphi(x)[/tex]
    [tex]\exists x\in S,\neg\varphi(x)[/tex]
    [tex]\forall x\in S,\neg\varphi(x)[/tex]

    The negation of "all cats are mammals" is "there is a cat that is not a mammal".
    Last edited: May 19, 2008
  9. May 19, 2008 #8
    The complement of the empty set (the real line in 1D, for example) is also both open and closed.

    For instance, any neighborhood of any point element of the real is also element of the real, so the set is open. Also, all accumulation points are elements of the real, so the set is closed.
  10. May 19, 2008 #9
    HA! Get it!

  11. May 19, 2008 #10


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    The empty set is open because the statement: "if x in A, some neighborhood of x is a subset of A" is true!

    If A is empty, the hypothesis: "if x in A" is false and so the statement is vacuously true.
    Last edited: May 19, 2008
  12. May 19, 2008 #11
    Closed is not the opposite of open, it is the dual. I.e., if a set A is open, you do not know that it is not closed, but you do know that the complement of A is closed. (same goes if we switch all instances of open and closed in that sentence)

    Someone said that it is not necessarily the case that the entire space is closed. However, when a set is open or closed, it is open or closed with respect to some set. As we have shown, the empty set is both open and closed with respect to any metric space. The complement of the empty set is the entire metric space, so this means that the entire metric space is both open and closed with respect to itself. Think about what this means for a while, and you'll see that it's obviously true.

    A comment that might help: For a sequence to have a limit the limit needs to lie in the metric space. For instance, if our metric space is [tex]\mathbb{Q}[/tex] and we have a sequence whose limit is [tex]\sqrt{2}[/tex] then the sequence does not have a limit because [tex]\sqrt{2} \notin \mathbb{Q}[/tex]. This means that the set [tex][0, \sqrt{2}) \cap \mathbb{Q}[/tex] is closed in [tex]\mathbb{Q}[/tex]
  13. May 19, 2008 #12
    Actually, that example was quite insightful. I am just reading Apostol's Methematical Analysis text and so far the setting is R-n. Of course, the definition of a closed set is given as a set whose complement in R-n is open, but I had not realized that the definition could be extended to the complement in any set and that a set could be open wrt a given set and closed wrt another set.
  14. May 21, 2008 #13


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    The definition of "topological space" (as opposed to the narrower "metric space) is a set, X, together with a collection, T, of subsets of X (called a "topology" for X) such that:
    1) All unions of members of T are members of T
    2) All finite intersections of T are members of T
    3) X itself is in T
    4) The empty set is in T

    Of course, the members of T are called "open" sets. If the empty set weren't open in a "metric topology", it wouldn't be a topology!
  15. May 21, 2008 #14
    Exactly. You can prove that the open sets of a metric space form a topology, and the question posed in this topic was about part of that proof. (Of course, the definition of a topological space in general is often not presented in your first exposure to analysis)
  16. Jan 12, 2010 #15
    This has always bothered me, why does it mean that the implication is true? I am more inclined to say the implication makes no sense, doesn't apply, is undefined, in our case. You could define the result of that implication to be true, but why would it follow from 'sense' that it is true? In other words, how do you know that the implication actually is true? Can't I say, if the moon is made of cheese, then books are made of hair?

    How is that implication true?

    Also, if you can merely define it to be true then you could define it to be false, and now what?
    Last edited: Jan 12, 2010
  17. Jan 12, 2010 #16
  18. Jan 12, 2010 #17
    The particular implication used in Mathematics is called "material implication" and it's defined by a truth table in which, for the cases where the antecedent is false, the implication is true. This may seem (and it is) at odds with our use of conditional sentences "If...then...", because we give them, consciously or not, a causal meaning; as in Mathematics, notions like causality and modality don't have significant roles, this type of implication works quite well, despite the occasional feeling of oddness (it occurs particularly in these limit cases that "stretch" the definitions).

    But material implication is only one of a class of logical entities known as Conditionals, and there are others that are more intuitive; the downside is that we don't have a clear and developed semantics for them.
  19. Jan 13, 2010 #18
    Never heard of it as an 'entity', checking it out now.

    Edit: quote from wikipedia

    'They are true because a material conditional is defined to be true when the antecedent is false (or the conclusion is true).'

    Looked up material conditional and it seems that there are two conditionals, logical and material. Logical seems to follow my intuition, where the answer is undefined if the antecedent is false (because it doesn't allow F antecedents). Material follows the conventional assignment of T to the implication when the antecedent is false.

    I still have seen no motivation for picking T over F in the assignment of the value of the material implication when the antecedent is false. (or a reason to pick either of them, for that matter)

    To pick out another non-sensical (to me) conditional: If the moon is made entirely of cheese, then the moon is made entirely of hair.

    And that statement is true?

    Edit 2: Didn't see that last post. I am not claiming any causality between them. I am familiar with such things as "If the ocean is blue, then a car has an engine." That seems vacuously true to me, where it is known that each is true and you just tie them together with an implication and call it good. It only seems to make sense to analyze the situation when you have variables in the conditional, say

    [tex](x>0) \rightarrow (y\le 5)[/tex]

    But anytime x is less than zero, somehow the implication is chosen to be true. It seems very arbitrary. Why can't I just pick a vacuous fallacy?

    Edit 3: Since there is a logical equivalent using only ands/ors/nots, why not just leave out the entire idea of implication? Why bother with it if it has an already well defined equivalent?
    Last edited: Jan 13, 2010
  20. Jan 13, 2010 #19
    The motivation to pick a definite truth value lies in the principle of bivalence, that states that all meaningful statements must be either true or false.

    The choice of the "true" truth value when the antecedent is false is because a number of reasons, for example:

    (1) When is a statement "If A then B" clearly false? Intuitively, when A is true and B is false; but this is the expression [tex]A \wedge \neg B[/tex]; so, the negation of this expression must be true; but the de Morgan's laws then state that [tex]\neg A\vee B[/tex] must be true, and is this last expression that it's used as a classical equivalent to the material implication.

    (2) Another reason is quantified statements like this one: "All men are mortal". Its logical form is:

    [tex]\forall x\left(A\left(x\right)\rightarrow B\left(x\right)\right)[/tex]

    But suppose you are quantifying over a domain larger than that of men; you still want that expression to be true, but the antecedent will be false for some elements of the enlarged domain, so it makes sense to consider the whole implication true.

    Regarding your example, if you interpret the If...then as the material conditional, then you have no other choice but to consider it true; but remember that, in natural language, the material implication is not always the best choice, and in that specific case, called a "counterfactual" is particularly bad.
    Suppose that, in some possible world, the moon is indeed a large hunk of cheese; then, in that world, the statement is false. Consider now a stronger form of implication that is only true if it's true in every possible world: with this implication, your statement would be false (but this is a case of a kind of modality, and mathematics is not modal, so material implication works well).

    Check the following links, if you are interested:



    EDIT: just saw your last post. You know why that sentence seems vacuously true? Because the consequent is very likely to be true. Those are easier to accept. It's also true (I wrote some of if above) that implication may be classically defined as [tex]\neg A \vee B[/tex] (but not, for example, if you want to give it a constructivist meaning, but that's another story); the reason it's not eliminated is because the If...then... construct seems to be very basic to our reasoning and inference processes; it would be very awkward to reason only in terms of "or" and "not".
    Last edited: Jan 13, 2010
  21. Jan 13, 2010 #20


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    I know it is OT, but I just started to wonder if I trust more mathematicians or lawyers.
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