# Convergence of a sequence in a metric space

## 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!

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

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.