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Question on Topological Spaces

  1. Feb 26, 2004 #1
    I'm a noob starting out studying differential geometry and topology. Really probably somewhere in the multivariate calculus level, but I've been trying to understand the plethora of terminology I'm encountering with this higher math. I've been reading a lot on Wikipedia.org and PlanetMath.org, (two excellent sites by the way)and I've got a question about topological spaces.

    From Wikipedia:

    The first requirement is that X is in T. If T is a subset of X, how can X be in T? Isn't that a paradox, sort of like Russell's paradox? How can a set be contained within a subset of itself? Am I interpreting the axiom wrong?

    Also, in reading about spaces in general, it seems like the general concept of space, metric spaces, euclidean space, etc .. are geometrical concepts. Objects like sets, groups, rings and fields are more algebraic in nature. However, the definition of a topological space seems to be more of an algebraic rather than a geometrical concept. Is that correct?
  2. jcsd
  3. Feb 26, 2004 #2
    T is a subset of the power set of X. the power set of X is the set of all subsets.

    and remember the distinction between subsets and proper subsets. any set is a subset of itself (this is why the symbol for subset is [itex]\subseteq[/itex], instead of just [itex]\subset[/itex]. this means that [itex]A\subseteq B[/itex] if A is equal to B or properly contained in B.

    so the power set of X contains all subsets of X, including X itself, and the empty set (which is a subset of every set).

    for example, if X={1,2,3}, then the power set of X is {0,{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}} (where 0 is the empty set). as you can see, X is itself a member of the power set of X. a topology is a subset of this power set that must contain X, 0, unions, and finite intersections. in particular, the power set is a subset of itself (but not a proper subset) which satisfies all these rules, so the power set of X is itself a topology of X, called the discrete topology

    yes, but what makes a metric space a geometrical concept is the presence of a metric. this is the distinction between geometry and topology; geometrical objects have a precise notion of distance, and topology have only a notion of closeness, without regard for absolute size, angle, curvature, etc.

    i would say that sets are not algebraic. set theory is the basis for the foundations of all mathematics, including analysis, algebra, geometry, and topology. so the definition of any object, be it topological, algebraic, analytic, or geometric, is a set theoretic definition.

    groups, rings, and fields are certainly algebraic objects. the property that these all have, and generic set theoretic objects do not, is a concept of multiplication or addition. anything with multiplication or addition is an algebraic object, and anything without is not.

    you might want to consider unions and intersections of sets as a kind of multiplication and addition. you can do this, but it is a more generic type of algebra that we usually consider for algebraic objects, so we don't usually consider these algebraic objects (although, sometimes we do)

    given my above comments, i would say that the definition of a topological space is a set theoretic definition. it relies on no algebraic properties of addition or multiplication, but only set theoretic notions like union and intersection.

    certainly, it is not geometric either, since it makes no mention of anything that could be considered a metric.

    the definitions of a topological space and a geometrical space (say, a metric space), are fairly similar; they are both set theoretic definitions. but then, as i said, all objects in mathematics have set theoretic definitions. the difference between the two is the presence of something that we call distance. in the case of the metric space, that notion is the metric itself.
  4. Feb 26, 2004 #3
    Ah, yes you are right, my sloppy reading of the definition of a subset. The very next sentence after my prior quote

    Here's another idea I'm struggling with though. The distinction between a closed set and an open set.

    If I'm not mistaken, an open set of the real numbers over the interval (0,1) would be open if:

    0 < x < 1

    but closed if:

    0 < x [tex]\leq[/tex] 1

    In either case you will never go beyond 1, and there is an infinite number of points along the interval in both cases. So what's the difference? Could it be that if x [tex]\leq[/tex] 1, then the value 1 is a solid final boundary, whereas if x < 1, and you are near, but not quite to 1, you can always advance closer to 1 over an infinite number of points. If x [tex]\leq[/tex] 1 however, once you are on the value one, you can't move any more. Is that on base at all?
    Last edited: Feb 26, 2004
  5. Feb 26, 2004 #4
    the definition is surprisingly simple in point set topology: a set is open if it is in the topology. it is essentially up to you which sets you want to call open, as long as they follow the rules.

    and a set is closed if its completent is open.

    in abstract algebra, you take the notion of multiplication, and define it to be any operation you want, as long as it follows a few rules. well, in point set topology, we do the same thing with open sets. define them to be any collection of sets which follow a few rules.

    the set (0,1) is an open subset of the real line with its usual topology, this is true, independent of what x is. (what is x, here?).

    um, even if x=100, (0,1) is still an open set. perhaps you want to talk about the set (0,1]? this set is neither open nor closed in the usual topology of the real line.

    it is not entirely clear what topological space you are in. i assumed the real line. is this what you had in mind?

    the difference between (0,1) and [0,1] is that surrounding any point in (0,1) there is a neighborhood contained within (0,1). this is sufficient to show that (0,1) is open. [0,1] does not satisfy this property (for example, no neighborhood of the point {1} is contained within [0,1]), so [0,1] is not open.

    i am not sure if that directed your question directly, but maybe it helped somewhat.
  6. Feb 26, 2004 #5
    Why must there be only open sets in a topology? Is that so that the rules that apply for sets in the middle also apply for sets at the edges?
  7. Feb 26, 2004 #6
    there mustn't only be open sets in topology, it's just that there must at least be open sets.

    and the reason we need open sets is that continuity is defined in terms of how it relates open sets to open sets. topology is the study of continuous maps.

    if you like, you can ditch open sets in favor of closed sets; it's equivalent. or something else, perhaps.
  8. Feb 26, 2004 #7

    matt grime

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    Temporarily forget about open and closed in the usual sense of intervals.

    A topology on a set is a collection of subsets satisfying the rules you gave. Elements of the topology are formally declared to be open, it's just a name. A set is closed iff its complement is in T. It is perfectly possible for sets to be both open and closed, open and not closed, closed and not open, and neither open nor closed.


    R with the sets generated by (a,b) under arbitrary union and finite intersection.

    Then (a,b) is open and not closed

    [a,b] is closed and not open

    (a,b] is neither closed nor open.

    Let R equipped with the discrete topology - every subset is open, and thus as every set is the complement of some other set, every set is closed too.

    any metric space comes with a topology from the metric.

    define a topology on R by saying

    a set is closed iff its elements are all roots of some finite number of polynomials - that is every closed set is just a finite number of points, every open set is the complement of this . this is the zariski topology.

    there is another interesting topology to do with arithmetic progressions but I can't remember it exactly at the moment that shows there are infinitely many primes.

    you need to stop thinking about edges and just go with the definitions. everything in maths is what it does, a simple philospophy that is necessary to understand anything at higher levels, so open sets are just things that satisfy certain rules - sometimes you might prove they satisfy these rules by the basic ideas you've seen already, but it won't help you understand the weak * topology on a Banach space.
  9. Feb 26, 2004 #8
    lethe, yes your answer helped.


    Is that how we manage to understand four dimensional space-time that we can't really visualize geometrically, but because we know the mathematical rules that apply at lower "visualizable" dimensions, we can simply follow the rules to apply them at higher "non-visualizable" dimensions?
  10. Feb 26, 2004 #9


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    You can do a lot that way, but the study of what really goes on in either Euclidean four dimensional space or in Minkowski four dimensional space is an active research topic. Papers are being published "as we speak".

    In the Euclidean context, we have the Geometrization Conjecture, which proposes to classify the three dimensional manifolds (analogs of the surfaces in 3-D) into a small number of specific types. If that conjecture is proved (it may already have been!) then it will imply that the Poincare conjecture is also true, which will solve one of the great problems of 20th century math.
  11. Feb 27, 2004 #10

    matt grime

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    It is the only way you can do maths at all after a certain point.

    In the case of topological spaces though, I meant that the ideas about edges, whilst suggestive and true in certain cases are not true in general - you need to define edge for abstract sets and topological spaces, the best alternative is 'boundary'. So the idea that [0,1] is closed becuase '1 is a solid boundary' 'you can't go beyond' is an idea that, if put more rigorously would be true for R in the usual topology (and for a metric toplogy), but doesn't make much sense for, say the Zariski topology topology on R ( a set is closed iff it contains a finite number of points) and in particular
    one space can have several different toplogies on it and some subset might be open in one topology and closed in another eg (0,1) in the usual and the discrete toplogy on R.
  12. Feb 27, 2004 #11
    Ok, this occurred to me last night about my first question. Back to the definition of a topological space, if it is defined as a collection of subsets T of the set X, and X is a member of the collection of sets, doesn't that define the entire set X?
    Last edited: Feb 27, 2004
  13. Feb 27, 2004 #12

    matt grime

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    Could you check that you've got all the letters in that question correct, please. What do you mean define the entire set X? It is required that the whole space thought of as a subset of itself, and the empty set are in the collection of sets defining the topology, that the whole space is open and closed, that's all. Other wise you'd have points that might not lie in any open (or closed) subset.

    A topological space is a set/space with a topology (collection of subsets satisfying certain rules), what in that is the thing you need to understand.

    A topological space is NOT defined to be the topology, it is a space equipped with a topology.
  14. Feb 27, 2004 #13
    Here's the http://planetmath.org/encyclopedia/TopologicalSpace.html [Broken] I'm looking at.

    Like I said, I'm a noob at this, trying to digest all of this new terminology, maybe I'm making it harder than it is.

    Here's another definition
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  15. Feb 27, 2004 #14

    matt grime

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    Yes, but can you please explain your question, cos I couldn't understand what you meant by 'defining X'

    Why is the fact that by defintion X is in T confusing you here?
  16. Feb 27, 2004 #15
    The way I understand it, a topological space T, is defined as a subset of a set X, or a topology on X. Is this definition meant to include the entire set X, or only a specific set contained within X, who's elements are less than the total elements contained in X (i.e., a specific "part" of X)?

    So to try to answer my own question, maybe a topological space could encompass the whole set T and maybe not. The topology is a set of subsets of a set X, satisfying specific properties, which resides within X, and that set X is the "space" in which the topology resides. Am I getting warmer?

    In fact, let me try an analogy:

    in [tex]R^2[/tex] Euclidean space with a traditional x,y Cartesian graph, you have sets of points which are defined by a distance function between them, and this distance function is a Metric on the space, i.e., the set of points defined by the Metric resides within the Euclidean space, sort of the way the T subsets satisfying the properties are a topology on a topological space X.
    Last edited: Feb 27, 2004
  17. Feb 27, 2004 #16

    matt grime

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    No. Is the shortest answer.

    Let us fix notation. Let X be a set. A topology on X is a collection of subsets, T, of X satisfying,

    for all indexing sets J, the union of elements of T (elements of T are subsets of X remember) indexed by J is also a member of T

    all finitely indexed intersections of elements in T are also in T

    the empty set is in T as is X.

    we say (X,T) is a topological space. Sometimes we suppress meniton of the T and call X a topological space. T is a topology. T is not a topological space, they are different.

    I don't know wher you've got the other things from, but they are wrong, as far as I can tell.

    Going all the way back to you first post, X has consistently been the topological space with topology T.

    "The way I understand it, a topological space T, is defined as a subset of a set X, or a topology on X"

    neither of those is correct. Your problems seem to be that you don't understand which thing is which.

    I don't follow most of the last paragraph:

    "maybe a topological space could encompass the whole set T and maybe not"

    what does encompass mean here? T is the topology, a collection of subsets of X, the topology does not "reside within" X in any way I can see - the elements of the topology are subsets of the set X, but that isn't the same thing.

    Here are examples of topologies, (X,T) would be the topological space.

    Let X be any set, there is the trivial topology T={ E,X} where I use E for the empty set.

    There is the discrete topology: T=P(X)

    There is a topology T={X subset of X : |X^c| is finite} u{E}--- a set is open iff and only if its complement contains a finite number of points, or it is the empty set.

    If X is infinite then these are all different, so one can make a set into a topological space in many different ways. That is why one refers properly to (X,T) as a topologcial space, unless it's clear what the topology is.
  18. Feb 27, 2004 #17
    T is a subset of the power set of X, not of X itself.
  19. Feb 27, 2004 #18
    I mistyped above, I meant to say that the set X is the topological space. I don't think I'm that far off, I realize now that X is the topological space which has the set T, a subset of the power set of X, as a topology on it.

    Forgive my non-rigorous, non-formal language, but I picture a subset T of a set X as "residing within" X. Maybe that's not the right description, but I often see terminology such as "for T IN X" which I would understand as, contained within, and element of, or residing within. Those phrases are all equivalent to me. Is that wrong?
  20. Feb 27, 2004 #19

    matt grime

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    I think all those statements are ok, but you are using T for two different things. T as a topology - a collection of subsets of X, and T as an element of the topology.

    It is ok, in a non-rigorous sense, to say that an element of the toplogy resides in the space, meaning it is a subset of it, but it is wrong to say that a topology resides in the set X in this sense, I think that just comes down to a reasonable sense of semantics.

    Just differentiate between the topology T on a set X, and an element of the topology. In your terms, each element of T resides in X, but T does not reside in X. More precisely every element of T is a subset of X, but the collection T is not a subset of X.
  21. Feb 27, 2004 #20

    matt grime

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    No, the points are not defined by a distance function at all. Euclidean 2-space exists without a metric.

    You can use the distance function to define the open sets in a topology on R^2, which agrees with the product topology on R^2 - that's an interesting elementary problem for you to prove.
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