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Basic questions on Nakahra's definition of topological space

  1. Jun 28, 2015 #1
    In chapter 2.3 in Nakahara's book, Geometry, Topology and Physics, the following definition of a topological space is given.

    Let [itex]X[/itex] be any set and [itex]T=\{U_i | i \in I\}[/itex] denote a certain collection of subsets of [itex]X[/itex]. The pair [itex](X,T)[/itex] is a topological space if [itex]T[/itex] satisfies the following requirements

    1.) [itex]\emptyset, X \in T[/itex]

    2.) If [itex]T[/itex] is any (maybe infinite) subcollection of [itex]I[/itex], the family [itex]\{U_j|j \in J \} [/itex] satisfies [itex]\cup_{j \in J} U_j \in T[/itex]

    3.) If K is any finite subcollection of [itex]I[/itex], the family [itex]\{U_k|k \in K \} [/itex] satisfies [itex]\cap_{k \in K} U_k \in T[/itex]

    I'm not very familiar with mathematical notation and conventions, so this is a little confusing to me. Presently I have a few questions:

    1.) What do [itex]I[/itex], [itex]J[/itex], and [itex]K[/itex] denote? I don't think they have been referenced before in the text. Do they just represent integers?

    2.) I'm a little confused by the first criterion in the definition. I thought [itex]T[/itex] is a collection of subsets of [itex]X[/itex]. Does the first criterion mean that the set [itex]X[/itex] is in [itex]T[/itex] or that a union of sets in [itex]T[/itex] should yield [itex]X[/itex]?

    3.) Why are the last two criterion not considered self evident from the definition of [itex]T[/itex]. For instance, for the second criterion if [itex]J<I[/itex] how can a union of [itex]U_j[/itex] be anything but within [itex]T[/itex]?

    Thanks.
     
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  3. Jun 28, 2015 #2

    Orodruin

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    1) I would assume I, J, and K are sets of integers labeling the open sets in the topological space.

    2) T is a collection of subsets of X. It does not mean that T contains all subsets of X. By definition, we call the sets which are in T open sets. The first criterion states that the empty set as well as the full set X are in T.

    3) No, these are not self evident. Not by a long shot. It is easy to find collections of sets which this does not hold for. For example, consider the set {1,2,3}. If we would let T contain the set itself, the empty set, {1}, and {2}, it would not satisfy these conditions. In particular, the union of the two last sets would not be in T.
     
  4. Jun 28, 2015 #3
    Ahh, it's not the elements we are concerned about, it's the sets themselves. So for example, {1,2,3} is an element of T, not 1, 2, etc. Thank you, this clarifies things significantly.
     
  5. Jun 28, 2015 #4

    mathman

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    1.) They are simply integers In this case there is a countable collection of subsets. In general the definition of topology is used even if the subset collection is not countable.
    2.) The statement simply means that X is a member of T.
    3.) They are not self evident. T can be any collection. Simple example. X has three point a,b,c. Let T consist of the following subsets {[itex]\phi[/itex]},{a},{b},{c}, {a,b,c}. This will not be a topology since {a}∪{b}={a,b} is not in T.

    To make things clear I am using {...} to mean a subset containing these elements.
     
  6. Jun 28, 2015 #5
    Thank you both for the responses. I feel like I understand now. I was confused about what T was.
     
  7. Jun 28, 2015 #6

    Orodruin

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    By the way, you should check whether the following Ts satisfy the conditions, it is a good exercise:

    1) The set T = {∅,X}.

    2) The set T = {U: U ⊆ X}, i.e., T is the set containing all subsets of X.
     
  8. Jun 30, 2015 #7
    I'm not really sure how to write this using proper mathematical notation and terminology so I will primarily use words, but please tell me if this is correct.

    It appears to me that both T satisfy the condition.

    1.) This satisfies condition 1 since T contains [itex]\emptyset[/itex] and X. This satisfies condition 2 since [itex]\emptyset \cup X = X \in T[/itex]. This satisfies condition 3 since [itex] \emptyset \cap X = \emptyset \in T [/itex].

    2.) This satisfies condition 1 since all possible subsets necessarily contain both the empty set and the set itself. This satisfies condition 2 since a union of subsets is always a subset and T contains all possible subset. This satisfies condition 3 since an intersection of subsets is either a subset or the empty set, both of which are contained in T.

    Is this correct?
     
  9. Jun 30, 2015 #8

    Orodruin

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    Yes. Both T are topologies on any space. They are called the trivial and discrete topologies, respectively. When encountering new topological concepts, I found it instructive to check the implications for these two topologies, e.g., checking what it means for a series to converge in a given topology.
     
  10. Jul 3, 2015 #9

    Fredrik

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    In notations like ##T=\{U_i|i\in I\}##, ##I## is called an index set. It's a set (any set) that can be bijectively mapped onto ##T##. The notation is useful e.g. when the elements of T are subsets of some set X, and you want to say something about a union of some of them. The union of all the elements of T can be denoted by ##\bigcup T##, but the index set gives you the option to write it as ##\bigcup_{i\in I}U_i## instead. If you want a notation for a union of some but not all of them, you can introduce a set ##J## that's a proper subset of ##I##, and write ##\bigcup_{i\in J}U_i##.
     
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