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Metric Spaces - Open subset

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex]a,b \in \mathbb{R}[/itex] and we define [itex]I = [a,b][/itex] ([] means closed set). Let [itex]\mathcal{C}_{\mathbb{K}}(I)[/itex] be the space of all continuous functions [itex]I \to \mathbb{K}[/itex] with the norm [itex]f \mapsto ||f||_I = \displaystyle \sup _{x \in I} |f(x)|[/itex]. Let [itex]U[/itex] be the set of all continuous functions [itex]I \to \mathbb{K}[/itex] so that [itex]\displaystyle |\int _a ^b f(x)dx| < 1[/itex]. Show that [itex]U[/itex] is an open subset of [itex]\mathcal{C}_{\mathbb{K}}(I)[/itex].


    2. Relevant equations
    If [itex]f: X \to Y[/itex] is a continuous function and [itex]B \subset Y[/itex] is open then the preimage [itex]f^{-1}[/itex] is open.

    For [itex]B \subset Y[/itex] the set [itex]f^{-1} = \{x \in X : f(x) \in B\}[/itex] is called the preimage of B.

    [itex]\mathbb{K} = \mathbb{R}[/itex] or [itex]\mathbb{C}[/itex]

    3. The attempt at a solution

    The function [itex]\mathcal{C} _{\mathbb{K}} (I) \to \mathbb{K}, f \mapsto \displaystyle \int _a ^b f(x)dx[/itex] is continuous because

    [itex] \displaystyle |\int _a ^b f(x)dx| \leq \int _a ^b |f(x)|dx \leq \int _a ^b ||f||_I dx = ||f||_I \int _a ^b dx = (b-a)||f||_I[/itex]

    I want to show that preimage of U is open then U is open using the two definitions I listed above. I'm just curious whether that is a legit method to solve this problem. I'm also wondering what the purpose of the norm is in this problem. If anyone can answer these questions I will be grateful.

    Thanks.
     
  2. jcsd
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