Checking if a set is open and relatively open in the sphere $\{x^2 + y^2 + z^2 = 1\}$

Question

Let S be the sphere $\{(x,y,z) \in \mathbb R^3 \mid x^2 + y^2 + z^2 = 1\}$.

First I want to check if the set $A = \{(\cos\phi \sin\theta, \sin\phi \sin\theta, \cos\theta ) \mid\phi \in (0,\pi), \theta \in (0,\pi)\}$ is open in $\mathbb R^3$.

I am pretty sure that it isn't open since it is half of the sphere S, and for the example for the point $(0,1,0) \in A$ we can't find a ball in $A$ that surrounds it.

The second task is to find whether $A$ is relatively open in $S$.

Here I am not so sure but I think I can take the open set: $V = $ $B(0,2)$ with $y>0$ and then $S \bigcap V = A$ which proves that it is relatively open in $S$.

However the set $B = \{ (\cos\phi \sin\theta, \sin\phi \sin\theta, \cos\theta ) \mid\phi \in [0,\pi], \theta \in [0,\pi]\}$ is not relatively open in $S$, right?

Am I correct here?

Help would be appreciated.

Answer 1

Suppose $U$ is an open set in $\mathbb{R}^3$ so that $U\cap S=B$. Then since $U$ is open, there is a ball $D$ around, say, the point $(0,0,1)$ with radius $\epsilon$ that is contained inside $U$. But since $U\cap S=B$ and $D\subset U$, it follows that $D\cap S\subset B$. In other words, the small ball $D$ centered at the north pole intersects the sphere in some open set $D\cap S$ which will be contained in the closed half-sphere $B$. But this is impossible, you can find points in the intersection $D\cap S$ which have $\theta's$ that fall outside of $0\leq\theta\leq \pi$.