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 Ted123 Nov25-11 04:43 PM

Convergence of a sequence in a metric space

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

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

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

3. The attempt at a solution

$x_n\to x$ in $(\mathbb{R},d_2) \iff d_2(x_n,x)\to 0$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ ;\;\;\;\;\;\;\;\;\, \iff |\tan^{-1}(x_n) - \tan^{-1}(x) | \to 0$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ ;\;\;\;\;\;\;\;\;\, \iff \tan^{-1}(x_n) \to \tan^{-1}(x)$

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

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ ;\;\;\;\;\;\;\;\;\, \iff |x_n - x| \to 0$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ ;\;\;\;\;\;\;\;\;\, \iff d_1(x_n,x) \to 0$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ ;\;\;\;\;\;\;\;\;\, \iff x_n\to x$ in $(\mathbb{R},d_1)$

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 Nov25-11 05:05 PM

Re: Convergence of a sequence in a metric space

Quote:
 Quote by Ted123 (Post 3634691) $\tan^{-1}(x_n) \to \tan^{-1}(x)\iff x_n \to x$ pointwise (since $\tan^{-1} : \mathbb{R} \to (-\pi/2 ,\pi/2)$ 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 $\lim_{n\to \infty} x_n = x$, then $\lim_{n \to \infty} \tan^{-1}(x_n) = \tan^{-1}(x)$ by the continuity of arctan, but the converse isn't always true for a given continuous function. For instance, take $f(x) = x^2$ and $x_n = -1$ for all n. Then $\lim_{n \to \infty} f(x_n) = 1 = f(1)$, but $\lim_{n \to \infty} x_n \neq 1$.

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

 Ted123 Nov26-11 05:16 AM

Re: Convergence of a sequence in a metric space

Quote:
 Quote by spamiam (Post 3634718) 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 $\lim_{n\to \infty} x_n = x$, then $\lim_{n \to \infty} \tan^{-1}(x_n) = \tan^{-1}(x)$ by the continuity of arctan, but the converse isn't always true for a given continuous function. For instance, take $f(x) = x^2$ and $x_n = -1$ for all n. Then $\lim_{n \to \infty} f(x_n) = 1 = f(1)$, but $\lim_{n \to \infty} x_n \neq 1$. 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 $d_2$ is a metric with the property that $d_2(x,y)< \pi$ for all $x,y\in\mathbb{R}$ - does this help? I can't see what property of arctan makes this implication true.

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