How can the inverse image of a continuous function not be closed?

Question

The inverse image of the set $\{x\in\mathbb{R}:x\geq 0\}$ through the continuous function \begin{align*} a:[0,1)&\to\mathbb{R}\\ s&\mapsto\cos(2\pi s) \end{align*} is $$a^{-1}([0,+\infty))=[0,1/4]\cup[3/4,1)$$ which is not closed in $\mathbb{R}$. I gather that the inverse image of this set is closed relative to $[0,1)$, but then my question is: are there any conditions that guarantee that $a^{-1}([0,+\infty))$ is closed in $\mathbb{R}$?

Answer 1

Yes: the inverse image of a continuous function $a\colon[0,1)\longrightarrow\mathbb R$ will be a closed subset of $\mathbb R$ if and only if $1$ does not belong to its closure in $\mathbb R$.

Answer 2

The set $a^{-1}([0, \infty))$ is closed in $\mathbb R$ if and only if it does not contain a "neighborhood of 1", that is, if there is some $x_0 < 1$ such that for all $x \in [x_0, 1)$, we have that $x \notin a^{-1}([0, \infty))$. Equivalently, if and only if there is an $x_0 \in [0, 1)$ such that the mapping $a$ maps the interval $[x_0, 1)$ into the negative numbers.