If $ (a_n)$ is a Cauchy sequence, then $ \arctan(a_n)$ is a Cauchy sequence.

Question

We know that if a function $f:A\mapsto\mathbb{R}$, $A\subseteq\mathbb{R}$, is uniformly continuous on $A$ then, if ($a_n$) is a Cauchy sequence in $A$, then ($f(a_n))$ is also a Cauchy sequence.

We know $\arctan$ is uniformly continuous on $A=\mathbb{R}$ however since the range of the function is $(-\frac{\pi}{2},\frac{\pi}{2})\subsetneq\mathbb{R}$ and the definition is a function from $A\rightarrow \mathbb{R}$, does this still apply? Intuitively, for this case, it seems so but I just want to make sure I'm not forgetting a possibility where it doesn't work.

Thank you.

Answer 1

Since $(a_{n})$ is Cauchy it converges to some value $\ell$. Arctan is a continuous function on the reals and at $\ell$ as consequence. Therefore by Cauchy criterion of continuouty $f(a_n)$ converges as well.

Answer 2

For functions $f : X \to Y$ the meaning of the word "range" is not standardized. Some people understand it is as the codomain $Y$ of $f$, other people understand it is a as the image $f(X) \subset Y$ of $f$. See here.

Let us come to the $\arctan$-function (principal value). We can understand it is a bijective function $\arctan : \mathbb R \to (-\pi/2,\pi/2)$ with codomain $(-\pi/2,\pi/2)$ or as an injective function $\arctan : \mathbb R \to \mathbb R$ with image $(-\pi/2,\pi/2)$. In my opinion there is no "correct" interpretation, it is a matter of taste.

You say that

• We know that if a function $f:A \to \mathbb R$, $A \subset \mathbb R$, is uniformly continuous on $A$, then if $(a_n)$ is a Cauchy sequence in $A$, then $(f(a_n))$ is also a Cauchy sequence.

Your doubt seems to be this:

You know that $\arctan$ is uniformly continuous on $A= \mathbb R$, but you regard it as a function with codomain $(-\pi/2,\pi/2)$ and not as a function with codomain $\mathbb R$. So why does the above bullet point apply here?

The uniform continuity of a real-valued function $f$ does not depend on the understanding of the codomain of $f$. In our case the metric on $(-\pi/2,\pi/2)$ is the restriction of the metric on $\mathbb R$. In both sets it is given by $\lvert a - b \rvert$. In other words, we may regard $\arctan$ as a function with codomain $\mathbb R$ so that the above bullet point applies.