Is the image of a dense set of a bijective continuous set dense?

Question

I have a dense set $K\subset (0,\pi)$. We have the continuous bijective function $h=\theta\mapsto\cot(\theta)$. This function $h:(0,\pi)\rightarrow \mathbb{R}$. My question is: is $h(K)$ dense in $\mathbb{R}$?

The main problem I have with this question is that $\cot$ maps values around $\frac{\pi}{2}$ to values around $0$. Points close to $0$ and $\pi$ are mapped to $\pm\infty$. This happens in such a way that the distance between $h(a)$ and $h(b)$ gets increasingly large as $a$ and $b$ get farther from $\frac{\pi}{2}$. So with values around $\frac{\pi}{2}$ I believe that the image of my dense set will be dense, but for values further from $\frac{\pi}{2}$ I am no sure.

Answer 1

Let $x \in \Bbb R$ be arbitrary. Consider $\theta = \cot^{-1}x \in (0, \pi).$

There exists a sequence $(k_n)$ of elements in $K$ such that $k_n \to \theta.$

Since $\cot$ is continuous, we have that $\cot k_n \to \cot\theta = x$.

Note that $(\cot k_n)$ is a sequence in $h(K)$.

Do you see how this answers your question?