Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question about proof from Bishop & Goldberg

  1. Feb 20, 2016 #1

    TeethWhitener

    User Avatar
    Science Advisor
    Gold Member

    I'm going through Bishop and Goldberg's "Tensor Analysis on Manifolds" right now and I'm stuck in Chapter 0. :H They give a proof of the statement "A compact subset of a Hausdorff space is closed" that I can't seem to wrap my head around. I'm reprinting the proof here:

    "Suppose that [itex]A[/itex] is a compact subset in a Hausdorff space [itex]X[/itex] and [itex]A \neq A^-[/itex] (where [itex]A^-[/itex] denotes the closure of [itex]A[/itex]), so there is an [itex]x \in A^- - A[/itex]. For every [itex]a \in A[/itex] there are open sets [itex]G_a, G^x_a[/itex] such that [itex]G_a \cap G^x_a = \emptyset , a \in G_a[/itex], and [itex]x \in G^x_a[/itex], because [itex]X[/itex] is Hausdorff. Then [itex]\{G_a|a \in A\}[/itex] is an open covering of [itex]A[/itex], so there is a finite subcovering [itex]\{G_a|a \in J\}[/itex], where [itex]J[/itex] is a finite subset of [itex]A[/itex]."

    I'm fine with everything up to this point, but the next sentence loses me:

    "But then [itex]\bigcap_{a \in J} G^x_a[/itex] is a neighborhood of [itex]x[/itex] which does not meet [itex]\bigcup_{a \in J} G_a \supset A[/itex], so [itex]x[/itex] cannot be in [itex]A^-[/itex], a contradiction."

    The authors have previously defined the closure of a set [itex]A[/itex] as the intersection of all closed sets containing [itex]A[/itex]. I get that [itex]\bigcap_{a \in J} G^x_a[/itex] is not a subset of [itex]A[/itex], but I don't understand why that implies that [itex]x \notin A^-[/itex].

    [EDIT]: I recalled another section earlier in the book saying that [itex]x \in A^-[/itex] iff every neighborhood of [itex]x[/itex] intersects [itex]A[/itex], which makes the last sentence in question trivial.
     
    Last edited: Feb 20, 2016
  2. jcsd
  3. Feb 20, 2016 #2

    fresh_42

    Staff: Mentor

    [itex]\bigcap_{a \in J} G^x_a[/itex] is an open neighborhood of [itex]x[/itex] and [itex](\bigcap_{a' \in J} G^x_{a'}) \cap (\bigcup_{a \in J} G_a) = \bigcup_{a,a' \in J} (G^x_{a'} \cap G_a) = ∅[/itex].
    If ##G^x_{a'} \cap G_a ≠ ∅## for some pair ##(a',a)## we can replace ##G^x_{a'}## by ## G^x_{a'} \cap G^x_{a}##. Remember ##J## being finite.
    Thus we have for any ##x∈X\setminus{A}## an open neighborhood ##\bigcap_{a \in J} G^x_{a}## that does not intersect with ##A##, i.e. any such ##x## is an inner point of ##x∈X\setminus{A}## which means ##x∈X\setminus{A}## is open and ##A## is closed.
     
  4. Feb 21, 2016 #3

    TeethWhitener

    User Avatar
    Science Advisor
    Gold Member

    Thanks for the reply @fresh_42. Maybe you can help me out with one further question. The proof above relies on the fact that, according to the book, "[itex]x \in A^-[/itex] iff every neighborhood of [itex]x[/itex] intersects [itex]A[/itex]." But I can only seem to show this for open neighborhoods.

    For instance, let [itex]x \in A^-[/itex] and assume that there's a neighborhood [itex]X_0[/itex] of [itex]x[/itex] such that [itex]X_0 \cap A = \emptyset[/itex]. If [itex]X_0[/itex] is open, then the complement [itex]X^c_0[/itex] is closed and [itex]A \subset X^c_0[/itex]. The fact that [itex]X^c_0[/itex] is closed implies that [itex]A^- \subset X^c_0[/itex] as well. But this means that [itex]X^c_0 \cap A^- = A^-, X_0 \cap A^- = \emptyset[/itex] and therefore [itex]x \notin A^-[/itex].

    This only works because we assumed that [itex]X_0[/itex] was open. Can we say something analogous about non-open neighborhoods? I get the impression from looking around that in general, "neighborhood" tends to refer to "open neighborhood," but in the book that I'm using, "neighborhood of [itex]x[/itex]" is just defined as any set for which [itex]x[/itex] lies in the interior, without regard to whether that set is open/closed/both/neither.

    But upon further reflection, I'm thinking that maybe this doesn't hold for non-open neighborhoods. If we consider [itex]\mathbb{R}[/itex] with the standard topology, then the singleton set [itex]\{x\}[/itex] is closed, as it's the complement of [itex](-\infty , x)\cup(x,\infty)[/itex]. If we take [itex]\{x\}[/itex] as a neighborhood of [itex]x[/itex], then it clearly can be in [itex]A^-[/itex] without intersecting [itex]A[/itex]. But by the same token, [itex]\{x\}[/itex] doesn't seem to have an interior, so maybe it doesn't count as a neighborhood of [itex]x[/itex]. Is this right?

    [EDIT]: Just found that I misspoke above. The book actually says "[itex]x \in A^-[/itex] iff every basis neighborhood of [itex]x[/itex] intersects [itex]A[/itex]." The paragraph before that mentions open sets as a basis of neighborhoods. I still can't really figure out the importance of a neighborhood versus a basis neighborhood, though.
     
  5. Feb 21, 2016 #4

    fresh_42

    Staff: Mentor

    What does this mean "in the book that I'm using, "neighborhood of [itex]x[/itex]" is just defined as any set for which [itex]x[/itex] lies in the interior"? Let's say ##x## is in the interior of ##X_0## then this means there is a open neighbourhood ##U_0## of ##x## such that ##U_0 ⊆ X_0##. So we usually may substitute ##X_0## by ##U_0## and thus getting an open neighborhood. If ##x## happens to be in ##\overline{X_0} \setminus X_0## then it will not work, e.g. ##\{1\} ∈ [0,1] ∩ [1,2]##.
    Your paragraph before the last is correct.

    I don't know the definition of a basis neighbourhood, maybe you can help me with this.
    Usually one examines (open) coveries containing a given set or the whole topological space.
    In case of compactness this means there is a finite set of open coveries which makes it far easier to intersect or build unions of all closed or all open sets.
     
  6. Feb 21, 2016 #5

    TeethWhitener

    User Avatar
    Science Advisor
    Gold Member

    Thank you, this finally cleared the logjam in my head. Just to be clear, it doesn't matter whether [itex]X_0[/itex] is open or not: As long as the interior of [itex]X_0[/itex] is nonempty, we can use it in the proof instead and it'll apply to the whole neighborhood as a superset of the interior. This is why singleton sets don't work: their interior is empty.

    As for a basis of neighborhoods, the book I'm using doesn't give a great definition. Any open set in a topology can be expressed as a union of basis sets, so I assume a basis of neighborhoods is a collection of neighborhoods such that any other neighborhood can be expressed as a union of the basis neighborhoods. At any rate, the basis is defined in terms of open sets, and coupled with your insight above, I think everything is clearer for me now. Thanks again for your help.
     
  7. Feb 21, 2016 #6

    fresh_42

    Staff: Mentor

    You're welcome!
    A topology is defined by its open sets. What you call basis I know as open cover (maybe coverage in English? but Wiki says cover).
     
  8. Feb 23, 2016 #7
    Yes, ##x \in \bar{A}## iff every basis element containing ##x## intersects ##A##. Note however that the topology itself is a basis (How can you write any open set as a union of elements in this basis?). So that automatically ##x \in \bar{A}## iff every neighborhood of ##x## intersects ##A##.

    Edit: Most authors use "neighborhood of ##x## to mean any open set ##U## which contains ##x##, while some others define it as any set which contains an open set which contains ##x##, so always check this in the book you're using. Usually very minor modifications in arguments/proofs are needed.
     
    Last edited: Feb 23, 2016
  9. Feb 23, 2016 #8
    No, both concepts are very different. What you describe can be defined as: A collection of open sets ##U_{ \alpha}## such that ##A \subseteq \union_{ \alpha} U_{ \alpha}## is an open covering of ##A##. What the OP is describing is a basis, which is a collection of open subsets ## \mathscr{B}## such that for any open set ##U## and any ##x \in U##, we can find ##B \in \mathscr{B}## such that ##x \in B \subseteq U##.
     
  10. Feb 23, 2016 #9

    TeethWhitener

    User Avatar
    Science Advisor
    Gold Member

    Thanks, I got the impression from my book that a basis and an open cover were quite different as well, and along the lines of what you said above.
     
  11. Feb 23, 2016 #10
    Referring to the original post: The final point is that you have obtained a finite covering {Ga, a in J} of the set A by open sets. And for each Ga of these finitely many open sets, there is another open set Gax containing x that is disjoint from Ga.

    And since the finite intersection of open sets is open, and since the union of the finitely many sets {Ga, a in J} contains A, this guarantees that the intersection of the sets Gax:

    V = a ∈ J Gax

    will be an open set containing x that is disjoint from A. (Since each Gax is disjoint from the corresponding Ga.) And that's a contradiction, since by initially assuming that A was not closed we found the point x not in A but whose every open neighborhood (including V) had to intersect A.
     
  12. Feb 23, 2016 #11
    Just as a note on this, the interior of the singleton ##\{x\} \subseteq X## is not always empty. What happens if you take the topological space ##(X, \tau)## where ##\tau = \mathcal{P}(X)##?

    And yes, both definitions of neightborhood give the same characterization of closure, namely: A point ##x## is in the closure of ##A## if and only if every neighborhood of ##x## has nonempty intersection with ##A##. So it doesnt matter wether you think of a neighborhood as an open set containing ##x## or as a set containing an open set containing ##x##, for *this* particular proof.

    I will prove it here: Suppose every open neighborhood of ##x## intersects ##A##. Let ##V## be a neighborhood of ##A##, then ##V## contains an open set ##U## containing ##x##, since ##U \cap A \neq \emptyset##, then ##V \cap A \neq \emptyset## because ##V## contains ##U##.
    Conversely, suppose every neighborhood of ##x## (open or not) intersects ##A##, then in particular all the open neighrborhoods intersect ##A##.
    Now you can easily prove that ##x \in \bar{A}## if and only if every open neighborhood of ##x## intersects ##A## and this will complete the argument that both definitions of neighborhood can be used interchangeably to characterize closure points.
     
  13. Feb 23, 2016 #12

    TeethWhitener

    User Avatar
    Science Advisor
    Gold Member

    That's true. The post you quote was referring to an earlier post where I specifically mentioned singletons in ##\mathbb{R}## with the standard topology.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Question about proof from Bishop & Goldberg
Loading...