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How is a discrete topology a 0-manifold?

  1. Jan 15, 2013 #1
    I am new to manifolds and I read the fact that any discrete space is a 0 dimensional manifold. I am having a hard time understanding why and feel this is very basic.

    So to be a manifold, each point in the space should have a neighborhood about it that is homeomorphic to R^n. (and n will tell you what dimension the manifold is). To start, what is 0-dimension Euclidean space? Would this be just the empty set?

    Now secondly I start to think about what a homeomorphism implies. It's a bijection between two topological spaces where each direction of the mapping is continuous. (so open sets in one map to open sets in the other). This is what makes it difficult for me to understand how a discrete space could be a manifold - how would you get a bijection between a discrete space and euclidean space? I can understand it in one direction by doing something trivial (like say, every set in the discrete space S maps to the empty set or all of R or something like that). But how do you get it in the other direction?
  2. jcsd
  3. Jan 15, 2013 #2


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    If you are using the definition of a topological manifold as a 2nd countable, Hausdorff, locally euclidean space then a topological space M is a 0 - manifold if and only if it is a countable discrete space. First, let M be a 0 - manifold (R^0 = {0} if that was your question btw). Let p be an element of M then there exists a homeomorphism f:U -> {0} where U is a neighborhood of p. f^-1({0}) must be an open singleton contained in U so f^-1({0}) = {p} is open in M. M has a countable basis and since each singleton in M is open, each is an element of the basis so M has countably many points thus M is a countable discrete space. Conversely, if M is a countable discrete space then M is Hausdorff and 2nd countable of course and for any p in M the map f:{p} -> {0} is a homeomorphism from a neighborhood of p to {0} so it is locally euclidean of dimension 0 thus a 0 - manifold.
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