Derived Sets In Metric Spaces Are They Always Closed?

In the fascinating realm of mathematical analysis, metric spaces provide a foundational framework for studying concepts like distance, convergence, and continuity. Within these spaces, the notion of a derived set holds significant importance. A derived set, denoted as $A'$, comprises all the limit points of a given set $A$. Understanding the properties of derived sets is crucial for grasping the topological structure of metric spaces.

Unveiling Derived Sets and Limit Points

Before we delve into the question of whether every derived set in a metric space is closed, it's essential to define the key concepts involved. Let's begin by defining a metric space. A metric space $(X, d)$ consists of a set $X$ equipped with a metric $d$, which is a function that assigns a non-negative real number to every pair of points in $X$, representing the distance between them. The metric $d$ must satisfy certain properties, such as non-negativity, symmetry, the triangle inequality, and the condition that the distance between two distinct points is strictly positive.

Now, let's define a limit point. Given a subset $A$ of a metric space $(X, d)$, a point $x$ in $X$ is called a limit point of $A$ if every open ball centered at $x$ contains at least one point of $A$ distinct from $x$ itself. In other words, no matter how small a neighborhood we consider around $x$, there will always be points from $A$ (other than $x$) within that neighborhood. The derived set $A'$ of $A$ is then defined as the set of all limit points of $A$.

To solidify our understanding, let's consider some examples. In the set of real numbers $\mathbb{R}$ with the usual metric, the derived set of the open interval $(0, 1)$ is the closed interval $[0, 1]$. This is because every point in $[0, 1]$ is a limit point of $(0, 1)$. For instance, consider the point 0. Any open interval centered at 0, no matter how small, will contain points from $(0, 1)$. Similarly, the derived set of the set $A = \{1/n : n \in \mathbb{N}\}$ is the set $\{0\}$, as 0 is the only limit point of this set.

Exploring Closed Sets in Metric Spaces

Now that we have a solid understanding of derived sets and limit points, let's turn our attention to closed sets. A subset $A$ of a metric space $(X, d)$ is said to be closed if it contains all of its limit points. Equivalently, a set is closed if its complement (the set of all points in $X$ that are not in $A$) is open. An open set is one in which every point has a neighborhood entirely contained within the set.

Consider some examples of closed sets. In the real numbers, any closed interval $[a, b]$ is a closed set. This is because it contains all its limit points, which are the points in the interval itself, including the endpoints $a$ and $b$. The set of integers $\mathbb{Z}$ is also a closed set in $\mathbb{R}$, as it contains all its limit points (which, in this case, are the integers themselves). The empty set $\emptyset$ and the entire metric space $X$ are both considered to be closed sets by definition.

To further clarify the concept of closed sets, let's consider some examples of sets that are not closed. The open interval $(0, 1)$ in $\mathbb{R}$ is not closed because it does not contain its limit points 0 and 1. Similarly, the half-open interval $[0, 1)$ is not closed because it does not contain its limit point 1. A set can fail to be closed if it's missing some of its boundary points.

The Key Question: Are All Derived Sets Closed?

Now, we arrive at the central question: Is it true that every derived set in a metric space is closed? This is a fundamental question in topology, and the answer, as we will see, is yes. To prove this, we need to show that if $A'$ is the derived set of a set $A$, then the derived set of $A'$, denoted as $(A')'$, is a subset of $A'$. In other words, we need to show that every limit point of the derived set $A'$ is also a member of the derived set $A'$.

To prove that the derived set $A'$ is closed, we will demonstrate that every limit point of $A'$ is also a limit point of $A$. Let $x$ be a limit point of $A'$. This means that every open ball $B(x, r)$ centered at $x$ with radius $r > 0$ contains a point $y$ in $A'$ distinct from $x$. Since $y$ is in $A'$, it is a limit point of $A$. Therefore, every open ball centered at $y$ contains a point in $A$ different from $y$.

Consider the open ball $B(x, r)$. Since $y$ is a limit point of $A'$, there exists a point $y \in A'$, $y \neq x$, such that $y \in B(x, r)$. Now, let $r_1 = r - d(x, y) > 0$. Since $y \in A'$, there exists a point $z \in A$, $z \neq y$, such that $z \in B(y, r_1)$. By the triangle inequality,

$$d(x, z) \leq d(x, y) + d(y, z) < d(x, y) + r_1 = d(x, y) + r - d(x, y) = r$$

Thus, $z \in B(x, r)$. Moreover, $z \neq x$. If $z = x$, then we would have $d(x, y) < r_1$, which is a contradiction. Hence, every open ball centered at $x$ contains a point $z$ in $A$ distinct from $x$. This implies that $x$ is a limit point of $A$, so $x \in A'$.

Therefore, we have shown that if $x$ is a limit point of $A'$, then $x$ is also a limit point of $A$. This means that $(A')' \subseteq A'$, which proves that the derived set $A'$ is closed. This result is a fundamental property of derived sets in metric spaces and has important implications in topology and analysis.

Implications and Applications

The fact that derived sets are closed has several important implications and applications in mathematics. For example, it plays a crucial role in the study of perfect sets. A set is said to be perfect if it is closed and equal to its derived set. Perfect sets have a rich mathematical structure and are essential in various areas of analysis and topology, such as the study of continuous functions and fractal geometry.

Moreover, the property of derived sets being closed is instrumental in proving other important theorems in topology. It is used in demonstrating the Bolzano-Weierstrass theorem, which states that every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence. This theorem is a cornerstone of real analysis and is used extensively in proving other results.

In summary, the derived set of any subset in a metric space possesses the significant property of being closed. This result not only enhances our understanding of the topological structure of metric spaces but also plays a pivotal role in various advanced mathematical concepts and theorems.

Conclusion

In conclusion, our exploration of derived sets in metric spaces has led us to a definitive answer: every derived set in a metric space is indeed closed. We began by defining key concepts such as metric spaces, limit points, and derived sets. We then explored the notion of closed sets and presented examples to solidify our understanding. The central question of whether derived sets are closed was addressed through a rigorous proof, demonstrating that the set of limit points of a derived set is contained within the derived set itself. This property has significant implications in topology and analysis, particularly in the study of perfect sets and the Bolzano-Weierstrass theorem. Understanding the properties of derived sets is crucial for a deeper appreciation of the structure and behavior of sets within metric spaces. The concepts discussed here form the bedrock for more advanced topics in real analysis and topology, highlighting the importance of a solid foundation in these fundamental ideas.

Therefore, the correct answer to the question "Every derived set in a metric space is:" is A. Closed.