Proof: An Open Sphere is an Open Set
Setup and Definitions
An open sphere (or open ball) in a metric space \((X, d)\) with center \(c \in X\) and radius \(R > 0\) is the set:
\( B(c, R) = \{x \in X \mid d(x, c) < R\} \)
A set is open if for every point within the set, there exists a smaller open sphere centered at that point which is entirely contained within the original set.
Our goal is to prove that \(B(c, R)\) satisfies this definition.
The Proof
Let \(y\) be an arbitrary point in the open sphere \(B(c, R)\). Our goal is to find a radius \(\epsilon > 0\) such that the new open sphere \(B(y, \epsilon)\) is completely contained within \(B(c, R)\).
- Analyze the position of \(y\).
Since \(y \in B(c, R)\), by definition we know its distance from the center \(c\) is strictly less than the radius \(R\): \[ d(y, c) < R \] - Define a new radius \(\epsilon\).
The distance from our point \(y\) to the boundary of the sphere is \(R – d(y, c)\). Let’s choose this value as our new radius. \[ \epsilon = R – d(y, c) \] Since \(d(y, c) < R\), this radius \(\epsilon\) is strictly positive (\(\epsilon > 0\)), as required. - Show containment using the Triangle Inequality.
Now, we must show that the entire new sphere \(B(y, \epsilon)\) is a subset of the original sphere \(B(c, R)\). To do this, let \(z\) be any arbitrary point in \(B(y, \epsilon)\). By definition of \(B(y, \epsilon)\), we have: \[ d(z, y) < \epsilon \] Our goal is to show that \(z \in B(c, R)\), which means we must prove that \(d(z, c) < R\). Using the triangle inequality, we can relate the distances between \(z\), \(y\), and \(c\): \[ d(z, c) \le d(z, y) + d(y, c) \] Since we know \(d(z, y) < \epsilon\), we can strengthen the inequality: \[ d(z, c) < \epsilon + d(y, c) \] Now, we substitute our definition of \(\epsilon = R - d(y, c)\): \[ d(z, c) < (R - d(y, c)) + d(y, c) \] The \(d(y, c)\) terms cancel, leaving us with: \[ d(z, c) < R \] This is precisely the condition for \(z\) to be an element of \(B(c, R)\).
Since \(z\) was an arbitrary point in \(B(y, \epsilon)\), we have shown that \(B(y, \epsilon) \subseteq B(c, R)\). Furthermore, since \(y\) was an arbitrary point in \(B(c, R)\), we have shown that for any point in the sphere, there exists another open sphere around it that is fully contained within the original. This satisfies the definition of an open set.