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I Metrics which generate topologies

  1. Jan 11, 2017 #1
    Given a topological space ##(\chi, \tau)##, do mathematicians study the set of all metric functions ##d: \chi\times\chi \rightarrow [0,\infty)## that generate the topology ##\tau##? Maybe they would endow this set with additional structure too. Are there resources on this?

    Thanks
     
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  3. Jan 11, 2017 #2

    andrewkirk

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    Yes they do. We say the topological space is 'metrizable' if there exists a function of the type you describe that has the properties of a metric and the metric topology that it generates is the same as ##\tau##. Studying metrizability is a key part of studying topology. Metrics provide additional structure because they provide a notion of distance, which does not exist in a bare topological space.

    A commonly-used resource on this is Munkres' popular textbook 'Topology, a first course' in which a significant part of the second chapter is devoted to metrizability.
     
  4. Jan 11, 2017 #3
    Yes, I've seen the metrizability concept before. But my question is not so much on the existence of a metric function which generates the topology, but given you already know it exists, for instance the Euclidean topology, is there some structure to the set of all such metric functions. For example, it seems this set is closed under addition. One can also ask about the cardinality of such set. Since metrizability focuses on existence, I'm not sure it will help.
     
  5. Jan 12, 2017 #4

    Stephen Tashi

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    - "such metric functions" meaning the class of all metric function that are equivalent to the Euclidean metric function.

    Is there is a name for an algebraic structure of things that are closed under addition and multiplication by positive scalars, but not under subtraction? (I don't know.)

    Suppose ##T: \mathbb{R}^n \rightarrow \mathbb{R}^n## is a 1-1 continuous (in the Euclidean topology) mapping of ##\mathbb{R}^n## onto itself. And let ##d(x,y )## be the Euclidean metric. Can we show ##m(x,y) = d(T(x),T(y))## is a metric that is topologically equivalent to the Euclidean metric?

    If that idea holds up then we can ask if the converse holds - if every metric ##m(x,y)## that is topologically equivalent to the Euclidean metric on ##\mathbb{R}^n## can be realized as ##d(T(x),T(y))## for some ##T(x,y)##.

    Things will be more interesting if the converse is false. If the converse is true then it looks like the study of "all metrics equivalent to the Euclidean metric" just amounts to the study of all 1-1 continuous mappings of ##\mathbb{R}^n## onto itself.
     
  6. Jan 12, 2017 #5
    Why a class though? Since such metrics are written as ##d: \mathbb{R}\times\mathbb{R} \rightarrow [0,\infty)##, which can be thought as a subset of ##\mathbb{R} \times \mathbb{R} \times \mathbb{R}##, so the set of all metrics equivalent to the Euclidean metric would be well defined in set theory as a subset of ##\mathcal{P}(\mathbb{R} \times \mathbb{R} \times \mathbb{R})##.

    These are very interesting questions. I will definitely be looking at them in more detail.
     
  7. Jan 14, 2017 #6

    FactChecker

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    Pay special attention to @Stephen Tashi 's point that the set is not closed under subtraction. That doesn't leave much in abstract algebra that can apply. Maybe category theory has some use here, but I don't know enough about that.
     
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