# Proof: open sphere is an open set

1. Apr 1, 2015

### nuuskur

1. The problem statement, all variables and given/known data
Prove that an open sphere in $\mathbb{R}^m$ is an open set.

2. Relevant equations

3. The attempt at a solution

To show that an open sphere is an open set, any point inside the sphere has to be an interior point:
Let us have a sphere $B(P_0, r), r > 0$, where $P_0$ is the centerpoint and r is the radius of the sphere and also an arbitrary point $P$ inside the sphere. Hence $\exists\varepsilon>0\colon B(P,\varepsilon)\subset B(P_0, r)$ and as we assumed the $B(P_0, r)$ to be open, then it does not contain its boundary.

The idea is to use the triangle inequality by taking another random point in a sphere around the already random point P in the large sphere: $S\in B(P, \varepsilon)$. The distance between two points $d(A,B)$ (Pythagorean)
$d(S, P_0)\leq d(S,P) + d(P,P_0) <\varepsilon + d(P,P_0)$. This is a shaky part - how can I justify this inequality is always correct? I can visualize it in my head, but is it accurate?

Per that assumption I would fix $\varepsilon\colon = r - d(P,P_0) > 0$ since $P$ is an interior point and its distance cannot be greater than or equal to the radius of the large sphere, therefore $\varepsilon$ is always strictly positive and $\forall S\in B(P, \varepsilon_1)$ I can state that the distance $d(S,P_0) < \varepsilon + d(P,P_0) = r$ which would mean that the open sphere is an open set as all its points are interior points.

2. Apr 1, 2015

### PeroK

The argument is essentially correct, but things are a bit mixed up and it doesn't flow. I suggest trying to tidy it up. I would start with something like this:

Let $B(P_0, r)$ be an open sphere in $\mathbb{R}^m$

Let $P \in B(P_0, r)$

$d(P_0, P) < r$ hence $\exists \epsilon > 0$ ...

3. Apr 1, 2015

### micromass

Staff Emeritus
A silly comment, but I don't like the term "open sphere". The terminology "open ball" is standard, because a sphere is usually the name given to the boundary of the ball, not something in the interior.

4. Apr 1, 2015

### nuuskur

I knew there was something off in the translation. I'll use ball to describe the set and sphere to describe its boundary, then.