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Homework Statement
Prove that an open sphere in [itex]\mathbb{R}^m[/itex] is an open set.
Homework Equations
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 [itex]B(P_0, r), r > 0[/itex], where [itex]P_0[/itex] is the centerpoint and r is the radius of the sphere and also an arbitrary point [itex]P[/itex] inside the sphere. Hence [itex]\exists\varepsilon>0\colon B(P,\varepsilon)\subset B(P_0, r) [/itex] and as we assumed the [itex]B(P_0, r)[/itex] 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: [itex]S\in B(P, \varepsilon)[/itex]. The distance between two points [itex]d(A,B)[/itex] (Pythagorean)
[itex]d(S, P_0)\leq d(S,P) + d(P,P_0) <\varepsilon + d(P,P_0)[/itex]. 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 [itex]\varepsilon\colon = r - d(P,P_0) > 0[/itex] since [itex]P[/itex] is an interior point and its distance cannot be greater than or equal to the radius of the large sphere, therefore [itex]\varepsilon[/itex] is always strictly positive and [itex]\forall S\in B(P, \varepsilon_1)[/itex] I can state that the distance [itex]d(S,P_0) < \varepsilon + d(P,P_0) = r[/itex] which would mean that the open sphere is an open set as all its points are interior points.