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

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.

• It seems OK.${}{}{}$ – Giuseppe Negro Mar 9 '19 at 18:02
• Thanks. But how can I formally prove that $B$ is not relatively open? – Gabi G Mar 9 '19 at 18:16
• You already did, I think it is enough. – Giuseppe Negro Mar 9 '19 at 19:20
• I proved that A is. In the comment I am talking about B – Gabi G Mar 9 '19 at 19:29

## 1 Answer

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 $$0\leq\theta\leq \pi$$.