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Closed set (metric spaces)

  1. Nov 7, 2011 #1
    1. The problem statement, all variables and given/known data

    If [itex]f:\mathbb{R}\to\mathbb{R}[/itex] and [itex]g:\mathbb{R}\to\mathbb{R}[/itex] are continuous functions show that:

    (a) the graph of [itex]f[/itex], [itex]\{(x,f(x)) : x\in\mathbb{R} \}[/itex] is a closed subset of [itex]\mathbb{R}^2[/itex].

    (b) [itex]\{ (x,f(x),g(x)) : x\in \mathbb{R} \}[/itex] is a closed subset of [itex]\mathbb{R}^3[/itex].

    3. The attempt at a solution

    I've done (a): the graph can be written as [itex]\{ (x,y) \in \mathbb{R}^2: y-f(x) = 0 \}[/itex] so we can use preimages:

    Considering the function [itex]F : \mathbb{R}^2 \to \mathbb{R}[/itex] defined [itex]F(x,y) = y-f(x)[/itex]; [itex]F[/itex] is continuous and the graph of [itex]f[/itex] is the preimage [itex]F^*(0)[/itex] and since [itex]\{0\}[/itex] is closed so is the graph.

    (b) must be similar but I can't see how to write the set in a form where I can use preimages immediately.

    The set can be written as:

    [itex]\{ (x,y,z)\in\mathbb{R}^3 : y = f(x) , z = g(x) \}[/itex]

    i.e. [itex]\{ (x,y,z)\in\mathbb{R}^3 : y - f(x) = g(x) - z = 0 \}[/itex]
  2. jcsd
  3. Nov 8, 2011 #2
    Hint: What must be the values of [itex]a,b\in\mathbb{R}[/itex] so that [tex]a^2+b^2=0?[/tex]
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