Convergence of a sequence in a metric space

[Ted123]

Homework Statement

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).

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]

\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]

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 realized 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.