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Boundary of the rationals

  1. Aug 5, 2011 #1
    I've read in several places that the boundary of the rational numbers is the empty set. I feel I must be misinterpreting the definition of a boundary, because this doesn't seem right to me.

    My understanding of the boundary of a set S is that it is the set of all elements which can be approached from both the inside and the outside. That is, the set of all r such that r is the limit of a sequence in S and also the limit of a sequence outside of S.

    We know of course that every real number is the limit of a sequence of rational numbers. We know also that every real number r is the limit of the constant sequence (r). So shouldn't the boundary of the rationals be the set of all irrational numbers?
     
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  3. Aug 5, 2011 #2

    HallsofIvy

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    What you say is correct up to the last sentence:
    Therefore the boundary of the rational numbers, as a set of real numbers with the usual topology, is the set of all real numbers, both rational and irrational.

    The boundary of the rational numbers, as a subset of the rational numbers with the usual topology, is empty. Perhaps that is what you saw? In any topology, the entire space has empty boundary.
     
  4. Aug 5, 2011 #3
    Okay, I think that makes perfect sense, but just to clarify:

    The discrepancy is that the the rationals can have multiple algebraic structures, and therefore multiple boundaries, depending on whether we consider them in isolation, or as a subset of the reals. (Or, presumably, as a subset of some other completion that I know less about, like the p-adic numbers).
     
  5. Aug 5, 2011 #4

    HallsofIvy

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    Yes, many of the topological properties of sets depend upon whether the set is a subset of some larger topology. Those that do not (compactness for example) are called "intrinsic".
     
  6. Aug 5, 2011 #5
    But I don't know if it would make sense to talk about, e.g., the boundary of the rationals a stand-alone space; I assume you always talk about the boundary of a subset A embedded in a space X; usually A is a subspace of X, I think.
     
  7. Aug 6, 2011 #6
    HallsofIvy already pointed out that the boundary of [itex]\mathbb{Q}[/itex] considered as a subset of [itex]\mathbb{R}[/itex] is all of [itex]\mathbb{R}[/itex]. I just wanted to expand on that a little.

    If [itex]X[/itex] is a topological space and [itex]A \subset X[/itex], the boundary of [itex]A[/itex] is the set of points with this property: each neighborhood of the point intersects both [itex]A[/itex] and [itex]X\setminus{A}[/itex].

    [itex][X\setminus{A}[/itex] is the set difference: the set of elements of [itex]X[/itex] that are not elements of [itex]A][/itex].

    This definition of boundary is equivalent to the one you gave. (Needs proof, of course).

    Now we can see that if [itex]A = X[/itex] then [itex]X\setminus{A}[/itex] is empty, so there can't be any elements in the boundary. That's why the boundary of [itex]\mathbb{Q}[/itex] in itself is empty.

    The more interesting case is when [itex]A[/itex] is a proper subset of [itex]X[/itex]; for example [itex]\mathbb{Q}[/itex] and [itex]\mathbb{R}[/itex]. If [itex]x[/itex] is a rational number, then any neighborhood about [itex]x[/itex] contains both rationals and irrationals. So all of [itex]\mathbb{Q}[/itex] is in the boundary. But if [itex]x[/itex] is an irrational number, it also has the property that each of its neighborhoods contain both rationals an irrationals. So the irrationals are in the boundary too.

    So, the boundary of [itex]\mathbb{Q}[/itex] is all of [itex]\mathbb{R}[/itex].

    Hope this helps.
     
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