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Homework Help: Continuity of a mapping in the uniform topology

  1. Sep 27, 2010 #1

    radou

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    1. The problem statement, all variables and given/known data

    Let (a1, a2, ...) and (b1, b2, ...) be sequences of real numbers, where ai > 0, for every i. Let the map h : Rω --> Rω be defined with h((x1, x2, ...)) = (a1x1 + b1, a2x2 + b2, ...). One needs to investigate under what conditions on the numbers ai and bi h is continuous, if Rω is given the uniform topology.

    3. The attempt at a solution

    Now, in a previous exercise, it was shown that, if Rω (the set of all infinite sequences of real numbers) is given the product topology, h is a homeomorphism from of Rω with itself.

    Further on, I know that the uniform topology is finer than the product topology.

    Let x be a point in Rω, and h(x) its image. Let V be a neighborhood of h(x) in the product topology. V can be written as a union of basis elements from the product topology. Now, since the uniform topology is finer than the product topology, for every y in V we can find a basis element B of the uniform topology which is contained in the basis element of the product topology which contains y. Hence V equals the union of the basis elements B in the uniform topology.

    Now, since h is continuous in the product topology, for V, there exists a neighborhood U of x such that h(U) is contained in V. Again, since the uniform topology is finer than the product topology, we can represent U as a union of basis elements from the uniform topology, and h(U) is still contained in V. Since this holds, we conclude that h is continuous in the uniform topology.

    There doesn't seem to be any condition on the ai and bi' other than ai > 0, as stated in the problem formulation.

    There's probably something wrong with my way of reasoning here, so please correct me if I'm wrong.
     
  2. jcsd
  3. Sep 27, 2010 #2

    radou

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    Actually, does this hold in general?

    "Given a set X, a function f : X --> X, and two topologies on X, T and T', if f is continuous with respect to the topology T, and if T' is finer than T, then f is continuous with respect to T', too. "

    Or, even more general?

    "Given a continuous function f : X -- Y, where Tx and Ty are the topologies on X and Y, respectively, and given topologies Tx' and Ty' on X and Y, which are finer than Tx and Ty, respectively, f : X --> Y is continuous when X is given the topology Tx' and Y the topology Ty'."
     
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