Locally Closed Sets: Equivalent Definitions In Topology

Hey guys! Today, we're diving deep into the fascinating world of general topology to unravel the mystery of locally closed sets. You might have stumbled upon a couple of definitions and wondered if they're just different ways of saying the same thing. Well, buckle up, because we're about to explore the equivalence of these definitions, making sure you're crystal clear on what it means for a set to be locally closed.

Understanding Locally Closed Sets

Before we jump into the nitty-gritty, let's get a solid grasp of what locally closed sets are all about. In simple terms, a subset $A$ of a topological space $X$ is considered locally closed if it can be expressed as the intersection of an open set and a closed set. This intuitive definition lays the groundwork for understanding the more formal definitions we'll be discussing.

The concept of locally closed sets pops up in various areas of mathematics, especially when dealing with manifolds, algebraic geometry, and functional analysis. Knowing the different ways to characterize these sets can be super helpful in proving theorems and solving problems. So, let's get started!

Definition 1: Neighborhood Condition

The first definition states that a subset $A$ of a topological space $X$ is locally closed if every point in $A$ has a neighborhood $V$ in $X$ such that the intersection of $A$ and $V$ (denoted as $A \cap V$) is closed in $V$. Let's break this down:

• Every point in A: This means we need to check every single point that belongs to the set $A$.
• Has a neighborhood V in X: For each point $x$ in $A$, we need to find an open set $V$ in the larger space $X$ that contains $x$. This $V$ is our neighborhood.
• A ∩ V is closed in V: The intersection of $A$ and $V$ gives us the part of $A$ that lies within the neighborhood $V$. We want this intersection to be a closed set relative to $V$. In other words, if we consider $V$ as its own topological space, then $A \cap V$ should contain all its limit points that are also in $V$.

To make this definition more concrete, consider an example. Let $X = \mathbb{R}$ with the usual Euclidean topology, and let $A = (0, 1]$. Notice that $A$ is neither open nor closed in $\mathbb{R}$. Now, let's check if it's locally closed according to our first definition. For any point $x \in (0, 1)$, we can choose a small open interval $V = (x - \epsilon, x + \epsilon)$, where $\epsilon > 0$ is small enough such that $V \subset (0, 1]$. Then, $A \cap V = V$, which is an open interval. However, for the point $1 \in A$, we can choose a neighborhood $V = (0.5, 1.5)$. Then, $A \cap V = (0.5, 1]$, which is closed in $V$. Thus, $A$ satisfies the first definition of being locally closed.

Definition 2: Intersection of Open and Closed Sets

The second definition offers a more direct characterization: a subset $A$ of a topological space $X$ is locally closed if it can be written as the intersection of an open set $U$ and a closed set $C$ in $X$. In other words, $A = U \cap C$, where $U$ is open and $C$ is closed in $X$.

This definition is often easier to visualize and work with. It tells us that if we can find an open set and a closed set whose intersection gives us our set $A$, then $A$ is locally closed. Let's revisit our previous example to see how this definition applies.

Again, consider $X = \mathbb{R}$ and $A = (0, 1]$. We want to find an open set $U$ and a closed set $C$ such that $A = U \cap C$. We can choose $U = (-1, 1)$ which is an open set in $\mathbb{R}$, and $C = (0, 1]$, which is not closed in $\mathbb{R}$. However, we can also express $A$ as $U = (0,2)$ and $C = (-1,1]$, where $U \cap C = (0,1]$. Therefore, $A$ can be represented as an intersection of an open and closed set.

Proving the Equivalence

Now comes the crucial part: demonstrating that these two definitions are equivalent. This means we need to show that if a set satisfies the first definition, it must also satisfy the second, and vice versa. Let's break down the proof into two parts.

Part 1: Definition 1 implies Definition 2

Suppose $A$ is a subset of $X$ that satisfies the first definition. This means for every $x \in A$, there exists a neighborhood $V_x$ in $X$ such that $A \cap V_x$ is closed in $V_x$. Our goal is to show that $A$ can be written as the intersection of an open set and a closed set in $X$.

Let's define $U = \bigcup_{x \in A} V_x$. Since $U$ is a union of open sets, it is itself an open set in $X$. Now, consider the set $U \cap A$. For each $x \in A$, we know that $x \in V_x$, so $x \in U$. Thus, $A \subseteq U$, and $U \cap A = A$.

Now, we want to show that $A$ is the intersection of $U$ and a closed set. For each $x \in A$, $A \cap V_x$ is closed in $V_x$, which means that $A \cap V_x = C_x \cap V_x$ for some closed set $C_x$ in $X$. Let $C = \bigcap_{x \in A} C_x$. Then $C$ is closed in $X$ since it's an intersection of closed sets.

We claim that $A = U \cap C$. We already know that $A \subseteq U$. Also, for any $x \in A$, $A \cap V_x \subseteq C_x$, so $A \subseteq C$. Thus, $A \subseteq U \cap C$. Conversely, if $y \in U \cap C$, then $y \in U$ and $y \in C$. Since $y \in U$, $y \in V_x$ for some $x \in A$. Since $y \in C$, $y \in C_x$ for all $x \in A$. In particular, $y \in C_x$ for the $x$ such that $y \in V_x$. Thus, $y \in C_x \cap V_x = A \cap V_x$, which implies $y \in A$. Therefore, $U \cap C \subseteq A$. Combining both inclusions, we have $A = U \cap C$, where $U$ is open and $C$ is closed in $X$. This shows that $A$ satisfies the second definition.

Part 2: Definition 2 implies Definition 1

Now, let's assume that $A$ satisfies the second definition, meaning $A = U \cap C$, where $U$ is open and $C$ is closed in $X$. We need to show that for every $x \in A$, there exists a neighborhood $V$ in $X$ such that $A \cap V$ is closed in $V$.

Let $x \in A$. Since $A = U \cap C$, we know that $x \in U$ and $x \in C$. Because $U$ is open, we can choose $V = U$ as our neighborhood of $x$. Now, we need to examine $A \cap V$. Since $V = U$, we have $A \cap V = A \cap U = (U \cap C) \cap U = U \cap C = A$.

Thus, $A \cap V = A = U \cap C = C \cap V$. Since $C$ is closed in $X$, $C \cap V$ is closed in $V$. Therefore, $A \cap V$ is closed in $V$. This shows that for every $x \in A$, there exists a neighborhood $V$ such that $A \cap V$ is closed in $V$, which means $A$ satisfies the first definition.

Conclusion

Alright, guys! We've successfully demonstrated that the two definitions of locally closed sets are indeed equivalent. Whether you prefer to think of a locally closed set as one where every point has a neighborhood in which the set is closed relative to that neighborhood, or as the intersection of an open set and a closed set, you can rest assured that you're talking about the same thing. This understanding not only deepens your grasp of general topology but also equips you with valuable tools for tackling more advanced mathematical concepts. Keep exploring, and happy problem-solving!