considered as a subspace of the plane $R^2$, and let $F : [0, 1) \rightarrow S^1$ be the map defined by $f (t) = (cos 2πt, sin 2πt)$.
The fact that f is bijective and continuous follows from familiar properties of the trigonometric functions. But the function $f ^{−1}$ is not continuous. The image under f of the open set $U = [0, 1/4)$ of the domain, for instance, is not open in $S^1$, for the point $p = f (0)$ lies in no open set V of$ R^2$ such that $V ∩ S^1 ⊂ f (U)$. See Figure 18.4.
My question is that: why point $p =f(0)$ lies in no open set $V$ of$ R^2$
Suppose that such $V$ exists, since $V$ is open, there exists an open ball $B(f(0),r)\subset V$, $B(f(0),1)\cap S^1$ contains an element of the form $(cos(-2\pi c),sin(-2\pi c))$, where $0<c<1/4$ which is not in $f([0,1/4)$ since $d(f(0),(cos(-2\pi c),sin(-2\pi c)))=\sqrt{(1-cos(2\pi c)^2+sin(2\pi c)^2}=\sqrt{2(1-cos(2\pi c)}<r$ if $c$ is small, $y$-coordinantes of the elements of $f([0,1))$ are positive, and the $y$-coordinate of $(cos(-2\pi c),sin(-2\pi c))$ is negative.