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Topology question(s)

  1. Apr 15, 2008 #1


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    I'm a physics student and I'm trying to work my way through Isham's Modern differential geometry for physicists. I guess the first question would be what you guys think of this book, does it cover all the necessary stuff (it's my preparation for general relativity)? Sadly I'm already having trouble with the introductory chapter on topology.. I'll post my questions in this topic.

    Isham starts out by defining filters and neighbourhood structures on a set X. A topological space is then a special kind of neighbourhood space. The proof of the following theorem is left up to the reader, but this reader needs some pointers..

    "A neighbourhood space (X,N) is a topological space if and only if each filter N(x) has a filter base consisting of open sets."

    When a filter base consists of open sets, what consequences does that have for the filter it generates?
  2. jcsd
  3. Apr 15, 2008 #2
    You don't have to study differential geometry to prepare for general relativity. The standard books used for both undergrad and grad courses develop the geometry that is needed to understand the theory. It's surprisingly not that much. When you want to study a more advanced treatment of some of the topics in relativity, that book will come in handy.
  4. Apr 16, 2008 #3


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    It would help us if you defined your terms. How are topological spaces defined? Filters? Filter bases? Open sets? etc. etc.
  5. Apr 16, 2008 #4


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    Ok, here are all the definitions that have been used:

    The power set P(X) of a set X is a lattice, with [tex]A \preceq B[/tex] defined as [tex]A \subset B[/tex]. Now a subset U of P(X) is an upper set if [tex]\forall a,b \in P(x) [/tex] with [tex]a \subset b[/tex], [tex]a \in U [/tex] implies [tex]b \in U [/tex].

    A filter F on X is a family of subsets of X such that F is algebraically closed under finite intersections, F is an upper family, and the empty set is not in F.

    A filter base D is a family of non-empty subsets of X such that if [tex]A,B \in D[/tex], then there exists [tex]C \in D[/tex] such that [tex]C \subset (A \cap B)[/tex]

    A neighbourhood structure N on X is an assignment of a filter N(x) to each [tex]x \in X[/tex], all of whose elements contain the point x. The pair (N,X) is then called a neighbourhood space.

    Given a neighbourhood space (N,X) and any set [tex]A \in X[/tex], a point x in X is a boundary point of A if every neighbourhood of x intersects both A and the complement of A (a neighbourhood of x is an element of N(x)). An open set is then a set that contains none of its boundary points.

    I suppose "each filter N(x) has a filter base" means that the filters N(x) are generated by those bases. In that case, here's the final definition:
    If D is a filter base, then [tex]\uparrow(D) := \{B \subset X| \exists A \in D \ such \ that \ A \subset B\} [/tex] is a filter, which is said to be generated by D.

    Edit: forgot one.. A topological space is a neighbourhood space (X,N) in which, for all x in X and for all M in N(x), there exists [tex]M_1 \in N(x)[/tex] such that, for all [tex]y \in M_1[/tex], [tex]M \in N(y)[/tex].
    Last edited: Apr 16, 2008
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