Closed set (metric spaces)

  • Thread starter Ted123
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Homework Statement



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].

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]
 

Answers and Replies

  • #2
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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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