Is $[0,2\pi)$ a closed set?

Question

If $\mathbb{S}^2=\{(x,y)\in\mathbb{R}^2 \mid x^2+y^2=1\}$ and $\phi:[0,2\pi)\rightarrow\mathbb{S^2}$ defined by $\phi(t)=(\cos(t),\sin(t))$ then $\phi$ is a bijective and continuous function.

Applying the following proposition to $\phi$

$$\text{$f$ is continuous if and only if the preimage of a closed set by $f$ is a closed set}$$

can I conclude that $\phi^{-1}(\mathbb{S^2})=[0,2\pi)$ is a closed set?

Answer 1

Yes, clearly $[0,2\pi)$ is closed (and open) in the space $[0,2\pi)$ which is the domain of your function. $X$ is always closed in $X$.

Answer 2

$[0,2\pi)$ is clearly not a closed subset of $\mathbb R$ with the usual topology, since it has a boundary point that is not one of its members, namely $2\pi.$

But it is closed as a subset of the space consisting only of $[0,2\pi).$ The point $2\pi$ does not exist within that space and so cannot be a boundary point within that space of any set within that space.

Look up the term "subspace topology." A space whose points are the members of $[0,2\pi)$ is a subspace of $\mathbb R$ if the subspace topology is assigned to it.