Homework Help: Convergence of a sequence in a metric space

Ted123 · Nov 25, 2011

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

For [itex]x,y \in\mathbb{R}[/itex] define a metric on [itex]\mathbb{R}[/itex] by [tex]d_2(x,y) = |\tan^{-1}(x) - \tan^{-1}(y) |[/tex] where [itex]\tan^{-1}[/itex] is the principal branch of the inverse tangent, i.e. [itex]\tan^{-1} : \mathbb{R} \to (-\pi/2 ,\pi/2)[/itex].

If [itex](x_n)_{n\in\mathbb{N}}[/itex] is a sequence in [itex]\mathbb{R}[/itex] and [itex]x\in\mathbb{R}[/itex], show that [itex]x_n \to x[/itex] as [itex]n\to\infty[/itex] in [itex](\mathbb{R} ,d_1)[/itex] where [itex]d_1[/itex] is the standard metric [itex]d_1(x,y)=|x-y|[/itex] if and only if [itex]x_n \to x[/itex] as [itex]n\to\infty[/itex] in [itex](\mathbb{R} ,d_2)[/itex].

3. The attempt at a solution

[itex]x_n\to x[/itex] in [itex](\mathbb{R},d_2) \iff d_2(x_n,x)\to 0[/itex]

[itex]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \iff |\tan^{-1}(x_n) - \tan^{-1}(x) | \to 0[/itex]

[itex]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \iff \tan^{-1}(x_n) \to \tan^{-1}(x)[/itex]

[itex]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \iff x_n \to x[/itex] pointwise (since [itex]\tan^{-1} : \mathbb{R} \to (-\pi/2 ,\pi/2)[/itex] is continuous)

[itex]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \iff |x_n - x| \to 0[/itex]

[itex]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \iff d_1(x_n,x) \to 0[/itex]

[itex]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \iff x_n\to x[/itex] in [itex](\mathbb{R},d_1)[/itex]

I'm not sure whether the pointwise bit in the middle is correct (as that seems to imply pointwise convergence and convergence in a metric is the same, when it isn't) and that is the crucial step!

spamiam · Nov 25, 2011

Ted123 said: ↑

[itex]\tan^{-1}(x_n) \to \tan^{-1}(x)\iff x_n \to x[/itex] pointwise (since [itex]\tan^{-1} : \mathbb{R} \to (-\pi/2 ,\pi/2)[/itex] is continuous)

I'm not sure if this addresses the pointwise vs. metric convergence you ask about, but there is a problem in your justification here. You're right that if [itex] \lim_{n\to \infty} x_n = x [/itex], then [itex] \lim_{n \to \infty} \tan^{-1}(x_n) = \tan^{-1}(x)[/itex] by the continuity of arctan, but the converse isn't always true for a given continuous function. For instance, take [itex] f(x) = x^2 [/itex] and [itex] x_n = -1 [/itex] for all n. Then [itex] \lim_{n \to \infty} f(x_n) = 1 = f(1) [/itex], but [itex] \lim_{n \to \infty} x_n \neq 1 [/itex].

But this problem doesn't arise for arctan. Can you think what property of arctan makes this implication true?

Ted123 · Nov 26, 2011

spamiam said: ↑

I'm not sure if this addresses the pointwise vs. metric convergence you ask about, but there is a problem in your justification here. You're right that if [itex] \lim_{n\to \infty} x_n = x [/itex], then [itex] \lim_{n \to \infty} \tan^{-1}(x_n) = \tan^{-1}(x)[/itex] by the continuity of arctan, but the converse isn't always true for a given continuous function. For instance, take [itex] f(x) = x^2 [/itex] and [itex] x_n = -1 [/itex] for all n. Then [itex] \lim_{n \to \infty} f(x_n) = 1 = f(1) [/itex], but [itex] \lim_{n \to \infty} x_n \neq 1 [/itex].

But this problem doesn't arise for arctan. Can you think what property of arctan makes this implication true?

I've just realised that the question points out that [itex]d_2[/itex] is a metric with the property that [itex]d_2(x,y)< \pi[/itex] for all [itex]x,y\in\mathbb{R}[/itex] - does this help? I can't see what property of arctan makes this implication true.

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Homework Help: Convergence of a sequence in a metric space

Ted123

spamiam

Ted123