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The notion of continuity applied to sets

  1. Oct 20, 2003 #1
    i'm trying to define what it would mean for a set to be continuous.

    what i'd like to say is that S is continuous if it is homeomorphic to [0,1], (0,1], or (0,1). (perhaps that's redundant already?)

    but i'm not sure if that captures all the sets i'd like to think of as continuous. my main dilema is whether or not [0,1]x[0,1] would be continuous. is [0,1]^2 homeomorphic to [0,1]? i know there is a continuous "space filling curve" that maps [0,1] onto [0,1]^2 but i'm not sure it has a continuous inverse. does it?

    if not, then the definition would be that S is continuous if it is homeomorphic to I^n, where I is some interval and n is a cardinal number. in this definition, the infinite dimensional hypercube [0,1]^[0,1] is included so that infinite dimensional manifolds could be classified as continuous or not.
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  3. Oct 20, 2003 #2


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    If you stick to subsets of R1 then it looks to me like your "continuous sets" are simply connected sets.

    There is no homeomorphism between [0,1]x[0,1] and [0,1]. For one thing, homeomorphic subsets of R<sup>n</sup> must have the same dimension.
  4. Oct 20, 2003 #3
    i would be talking about any sets with a topology, not just subsets of R. in R, the only connected sets are (potentially collapsed) intervals, right? so it's automatic that a connected subset of R would be continuous by the requirement that it be homeomorphic to an interval.

    is that really the case that dimension is a topological property? i didn't know that. that's good to know. i guess that means the space filling curve has no inverse that is continuous. i'm saddened by that, actually.

    yeah, then that being the case, i'd have to say that if S is a subset of T, where (T, t) is a topological space, then S is a continuous set of T if S is homeomorphic to I^n, where n is a set with positive cardinality (such as 1, 2, or even [0,1] for hilbert cube type-thingies) and I is an interval that is open, closed, or half open.

    yes or no: there is an uber-manifold in which all other manifolds can be embedded.
  5. Oct 20, 2003 #4


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    So your definition of "continuous" is homeomorphic to some cartesian product of closed real intervals. OK, and what is your proposed use for these? Aside from the compactness, they seem to be pretty simple and limited. Consider manifolds, locally homeomorphic to a cartesian product of open real intervls (with a matching condition on the intersections).
  6. Oct 20, 2003 #5
    I can be open, half-open, or closed. if closed, then S would be compact, yes. a continuous set is not necessarily compact.

    use? description. just another adjective to put in front of subsets of topological spaces. i don't know what kind of "theory" might come out of it. maybe within certain kinds of topological spaces, the subsets that are continuous will automatically divide into different classes based on further properties. i'm wondering if continuous sets are automatically metrizable, which would suggest that non-metrizable spaces are automatically discontinuous.

    then you can have piecewise continuous, which would probably mean each connected component is continuous. i don't know, maybe certain things can be said in general about maps from (piecewise) continuous sets...

    the actually problem i'm addressing is whether or not the power set of binary sequences under a nontrivial topology (i don't have a particular one in mind yet) is continuous.
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