Closed Sets and Accumulation Points

Let \( (X, d) \) be a metric space. We say that a subset \( Y \subseteq X \) is closed with respect to the metric \( d \) if its complement is open. In other words, \( Y \) is closed if \( \forall y \notin Y \), \( \exists r > 0 \) such that:

\[
B_r(y) \subseteq Y^c = X \setminus Y.
\]

This means that for any point outside of \( Y \), there exists an open ball around that point, which lies entirely within the complement of \( Y \).

Example in \( \mathbb{R} \):

Consider the metric space \( \mathbb{R} \) with the usual metric \( d(x, y) = |x – y| \). The set \( Y = [0, 1] \) is a closed subset of \( \mathbb{R} \). Its complement \( Y^c = (-\infty, 0) \cup (1, \infty) \) is open because it is a union of open intervals.

Example in \( \mathbb{R}^2 \):

Consider the set \( Y = \{ (x, y) \in \mathbb{R}^2 : x^2 + y^2 \leq 1 \} \), which is the closed unit disk in \( \mathbb{R}^2 \). Its complement \( Y^c = \{ (x, y) \in \mathbb{R}^2 : x^2 + y^2 > 1 \} \) is open because it consists of all points whose distance from the origin is greater than 1, forming an open set.

Example in Function Spaces:

Consider the metric space \( C[0,1] \), the set of continuous functions on the interval \([0,1]\), with the metric \( d(f, g) = \sup_{x \in [0,1]} |f(x) – g(x)| \). Let \( Y = \{ f \in C[0,1] : f(0) = 0 \} \). Is \(Y\) a closed set? This is a very difficult question to answer directly based on the definition of closed set. To put ourselves in a position to answer this question, we need to derive a different characterization.

Let \( E \subseteq X \). The following are equivalent:

• (a) \( E \) is closed.
• (b) If \( (x_n)_{n \in \mathbb{N}} \) is a sequence of points in \( E \) and \( \lim_{n \to \infty} x_n = x \) in the metric space \( X \), then \( x \in E \).

Proof

Part 1: Closedness implies convergence property.

Suppose \( E \) is closed. Let \( (x_n)_{n \in \mathbb{N}} \) be a sequence in \( E \) such that \( \lim_{n \to \infty} x_n = x \). Assume for contradiction that \( x \notin E \).

Since \( E^c \) is open and \( x \in E^c \), there exists \( r > 0 \) such that \( B_r(x) \subseteq E^c \). Because \( \lim_{n \to \infty} x_n = x \), there exists \( N \in \mathbb{N} \) such that for all \( n > N \), \( x_n \in B_r(x) \subseteq E^c \). This contradicts the fact that \( x_n \in E \) for all \( n \). Therefore, \( x \in E \), and thus \( E \) contains all its limit points.

Part 2: Convergence property implies closedness.

Suppose that whenever a sequence \( (x_n) \) in \( E \) converges to \( x \in X \), then \( x \in E \). Assume for contradiction that \( E \) is not closed. Then \( E^c \) is not open, so there exists \( x \in E^c \) such that for every \( r > 0 \), \( B_r(x) \not\subseteq E^c \). This means that every open ball around \( x \) contains points of \( E \).

Therefore, we can construct a sequence \( (x_n) \) in \( E \) such that \( x_n \to x \). By assumption, \( x \in E \), which contradicts \( x \in E^c \). Hence, \( E \) must be closed.

Examples

Example in \( \mathbb{R} \):

Consider \( E = [0, 1) \) in \( \mathbb{R} \) with the usual metric. The sequence \( x_n = 1 – \frac{1}{n} \) is in \( E \) for all \( n \), and \( x_n \to 1 \) as \( n \to \infty \). However, \( 1 \notin E \). Therefore, \( E \) is not closed because it does not contain the limit of a convergent sequence within it.

Example in \( \mathbb{R}^2 \):

Let \( E = \{ (x, y) \in \mathbb{R}^2 : x > 0 \} \), the right half-plane excluding the \( y \)-axis. Consider the sequence \( \left( x_n, y_n \right) = \left( \frac{1}{n}, 0 \right) \), which lies in \( E \) for all \( n \). The sequence converges to \( (0, 0) \), but \( (0, 0) \notin E \). Thus, \( E \) is not closed because it does not contain the limit of a convergent sequence within it.

Example in Function Spaces:

Consider the space \( C[0,1] \) of continuous functions on \([0,1]\) with the supremum metric \( d(f, g) = \sup_{x \in [0,1]} |f(x) – g(x)| \). Let \( E = \{ f \in C[0,1] : f(0) > 0 \} \). The sequence \( f_n(x) = \frac{1}{n} + x^2 \) is in \( E \) for all \( n \), but it converges uniformly to \( f(x) = x^2 \), which is not in \( E \) since \( f(0) = 0 \). Therefore, \( E \) is not closed as it does not contain the limit of a convergent sequence within it.

Example in \( \ell^2 \) Space:

Let \( E \) be the set of sequences \( x = (x_n) \in \ell^2 \) such that \( x_n = 0 \) for all but finitely many \( n \). This set is not closed in \( \ell^2 \) because sequences in \( E \) can converge to sequences not in \( E \). For example, consider the sequence \( x^{(k)} \) where \( x^{(k)}_n = \delta_{nk} \) (1 at position \( k \), 0 elsewhere). As \( k \to \infty \), \( x^{(k)} \) converges weakly to the zero sequence, which is in \( E \), but the closure of \( E \) contains all sequences in \( \ell^2 \), showing that \( E \) is not closed.

An accumulation point \( x \) of a set \( E \subseteq X \) is a point where every open ball centered at \( x \) intersects \( E \) in a non-trivial way. That is, for all \( r > 0 \):

\[
B_r(x) \cap (E \setminus \{x\}) \neq \emptyset.
\]

In other words, there exists a sequence \( (x_n)_{n \in \mathbb{N}} \) such that \( x_n \in E \) and \( \lim_{n \to \infty} x_n = x \).

Theorem

A set \( E \subseteq X \) is closed if and only if it contains all of its accumulation points.

Examples

Example in \( \mathbb{R} \):

Consider the set \( E = \left\{ \frac{1}{n} : n \in \mathbb{N} \right\} \) in \( \mathbb{R} \). The point \( x = 0 \) is an accumulation point of \( E \) since \( \lim_{n \to \infty} \frac{1}{n} = 0 \). However, \( 0 \notin E \). Therefore, \( E \) is not closed because it does not contain all its accumulation points. The closure of \( E \) is \( E \cup \{0\} \).

Example in \( \mathbb{R}^2 \):

Consider \( E = \left\{ \left( \frac{1}{n}, \frac{1}{n} \right) : n \in \mathbb{N} \right\} \subseteq \mathbb{R}^2 \). The point \( (0, 0) \) is an accumulation point of \( E \) because \( \lim_{n \to \infty} \left( \frac{1}{n}, \frac{1}{n} \right) = (0, 0) \). However, \( (0, 0) \notin E \). Thus, \( E \) is not closed as it does not contain all its accumulation points.

Example in Function Spaces:

In the space \( C[0,1] \) with the supremum metric, consider the set \( E = \{ f_n : n \in \mathbb{N} \} \), where \( f_n(x) = x^n \). Each \( f_n \) is continuous on \([0,1]\) and belongs to \( E \). The pointwise limit of \( f_n \) as \( n \to \infty \) is the function \( f(x) = 0 \) for \( x \in [0,1) \) and \( f(1) = 1 \), which is not continuous at \( x = 1 \) and thus not in \( C[0,1] \). Therefore, \( E \) is not closed.

Example in the Rational Numbers:

Consider \( E = \mathbb{Q} \cap [0,1] \subseteq \mathbb{R} \). Every real number in \([0,1]\) is an accumulation point of \( E \) because rationals are dense in \(\mathbb{R}\). However, \( E \) does not contain all these accumulation points (the irrational numbers in \([0,1]\)). Therefore, \( E \) is not closed in \(\mathbb{R}\), and its closure is the entire interval \([0,1]\).

A point \( x \) is called a boundary point of \( E \) if for every \( r > 0 \):

\[
B_r(x) \cap E \neq \emptyset \quad \text{and} \quad B_r(x) \cap E^c \neq \emptyset.
\]

This means that a boundary point lies on the edge of the set, where points in both \( E \) and its complement \( E^c \) are arbitrarily close to \( x \).

Example in \( \mathbb{R} \):

In \( \mathbb{R} \), consider \( E = (0,1) \). The points \( x = 0 \) and \( x = 1 \) are boundary points of \( E \) because any open interval centered at \( 0 \) or \( 1 \) contains points from \( E \) and from \( E^c \). Points within \( (0,1) \) are not boundary points since there exist open intervals around them contained entirely within \( E \).

Example in \( \mathbb{R}^2 \):

Consider \( E = \{ (x, y) \in \mathbb{R}^2 : x^2 + y^2 < 1 \} \), the open unit disk. Points on the circle \( x^2 + y^2 = 1 \) are boundary points of \( E \) because any open ball centered at such a point contains points from \( E \) and \( E^c \). Points strictly inside the disk are not boundary points.

Example in the Complex Plane:

Consider \( E = \{ z \in \mathbb{C} : |z| < 1 \} \), the open unit disk in the complex plane. Points on the circle \( |z| = 1 \) are boundary points of \( E \). Any open neighborhood around such a point contains complex numbers inside and outside the disk, meeting the criteria for boundary points.