A Detailed Proof: Why a Closed Sphere is a Closed Set

Proof: A Closed Sphere is a Closed Set

Setup and Definitions

A closed sphere (or closed ball) in a metric space \((X, d)\) with center \(c \in X\) and radius \(r \ge 0\) is the set:

\( S = \{x \in X \mid d(x, c) \le r\} \)

A set is closed if its complement is an open set. We will prove that the complement of \(S\), denoted \(S^c\), is open.

Proof by Complement

The complement of the closed sphere \(S\) is the set of all points whose distance from \(c\) is strictly greater than \(r\):

\( S^c = X \setminus S = \{x \in X \mid d(x, c) > r\} \)

To show that \(S^c\) is open, we must show that for any point \(y \in S^c\), there exists an open ball centered at \(y\) that is entirely contained within \(S^c\).

1. Choose an arbitrary point in the complement.
   Let \(y \in S^c\). By definition, its distance from the center \(c\) is \(d(y, c) > r\). Let’s define this distance as \(R = d(y, c)\).
2. Define a suitable radius for an open ball around \(y\).
   The “gap” between \(y\) and the sphere \(S\) is \(R – r\). This value is strictly positive. Let us choose our radius to be \(\epsilon = R – r > 0\).
3. Show the open ball is contained in the complement.
   Consider the open ball \(B(y, \epsilon) = \{z \in X \mid d(z, y) < \epsilon\}\). We must show that \(B(y, \epsilon) \subseteq S^c\). Let \(z\) be any point in \(B(y, \epsilon)\), which means \(d(z, y) < \epsilon\). We need to show that \(z \in S^c\), which is equivalent to showing \(d(z, c) > r\). By the triangle inequality, we know \(d(c, y) \le d(c, z) + d(z, y)\). Rearranging this (the reverse triangle inequality) gives: \[ d(c, z) \ge d(c, y) – d(z, y) \] We know that \(d(c, y) = R\) and \(d(z, y) < \epsilon\). Substituting these into the inequality yields: \[ d(c, z) > R – \epsilon \] By our choice of \(\epsilon = R – r\), we have: \[ d(c, z) > R – (R – r) = r \] Thus, \(d(c, z) > r\). This confirms that \(z \in S^c\).

Since \(z\) was an arbitrary point in \(B(y, \epsilon)\), we have shown that \(B(y, \epsilon) \subseteq S^c\). And since \(y\) was an arbitrary point in \(S^c\), we conclude that \(S^c\) is an open set.

Alternative Proof Using Sequences

An alternative definition of a closed set is one that contains all of its limit points. We show that any convergent sequence of points in \(S\) has its limit in \(S\).

1. Let \((x_n)_{n=1}^\infty\) be a sequence of points such that \(x_n \in S\) for all \(n \in \mathbb{N}\).
2. Assume this sequence converges to a limit \(x \in X\), i.e., \(\lim_{n \to \infty} x_n = x\).
3. Since \(x_n \in S\) for all \(n\), the definition of \(S\) implies that \(d(x_n, c) \le r\) for all \(n\).
4. The distance function \(d\) is continuous with respect to its arguments. Therefore, the function \(f: X \to \mathbb{R}\) defined by \(f(y) = d(y, c)\) is continuous.
5. By the properties of continuous functions, if \(x_n \to x\), then \(f(x_n) \to f(x)\). This means: \[ \lim_{n \to \infty} d(x_n, c) = d(x, c) \]
6. We have a sequence of real numbers, \(a_n = d(x_n, c)\), where \(a_n \le r\) for all \(n\). A fundamental theorem of real analysis states that if a sequence converges, its limit must be less than or equal to any upper bound of the sequence.
7. Therefore, we must have \(d(x, c) \le r\). This is precisely the condition for \(x\) to be in \(S\).

Since the limit \(x\) of any convergent sequence from \(S\) is also in \(S\), the set \(S\) is closed.