Union Of Open Sets Is Open

Union of Open Sets is Open: A Proof and its Implications

This article explores the fundamental topological concept that the union of open sets is always open. We'll provide a rigorous proof, discuss its significance, and illustrate it with examples. Understanding this property is crucial for grasping many core concepts in topology and analysis. This proof is essential for students of mathematics and anyone working with topological spaces.

What are Open Sets?

Before diving into the proof, let's define what we mean by an "open set." In a metric space (like the real numbers with the usual distance function), a set is considered open if for every point within the set, you can find a small ball (or interval) centered at that point that is entirely contained within the set. This intuitively means there's "space" around each point. More formally, a set U in a metric space (X, d) is open if for every x ∈ U, there exists an ε > 0 such that the open ball B(x, ε) = {y ∈ X : d(x, y) < ε} is a subset of U. This definition generalizes to more abstract topological spaces using neighborhoods instead of open balls, but the core idea remains the same.

The Theorem: Union of Open Sets is Open

Theorem: Let (X, τ) be a topological space. If {U_α}_α∈A is a collection (possibly infinite) of open sets in X, then their union, ∪_α∈A U_α, is also an open set in X.

Proof:

To prove this theorem, we need to show that for any point x in the union ∪_α∈A U_α, there exists an open neighborhood around x that is entirely contained within the union.

1. Choose a Point: Let x ∈ ∪_α∈A U_α. By the definition of a union, this means that x belongs to at least one of the sets U_α. Let's say x ∈ U_β for some β ∈ A.

2. Use the Openness of U_β: Since U_β is an open set, there exists an open neighborhood N_x of x such that x ∈ N_x ⊂ U_β. The existence of this neighborhood is guaranteed by the definition of an open set.

3. Neighborhood within the Union: Because N_x ⊂ U_β and U_β ⊂ ∪_α∈A U_α (since U_β is one of the sets in the union), we have N_x ⊂ ∪_α∈A U_α.

4. Conclusion: We've shown that for an arbitrary point x in the union ∪_α∈A U_α, there exists an open neighborhood N_x entirely contained within the union. Therefore, by definition, ∪_α∈A U_α is an open set. This completes the proof.

Significance and Examples

This theorem is fundamental because it ensures that the collection of open sets in a topological space forms a topology, meaning it satisfies specific axioms. These axioms are necessary for defining continuous functions and other crucial topological concepts.

Examples:

• Real Numbers: Consider the open intervals (0, 1) and (1, 2) in the real numbers. Both are open sets. Their union, (0, 1) ∪ (1, 2), is also an open set. Note that the union is not necessarily a single interval; it can be a disjoint union of open sets.

• Infinite Union: The union of infinitely many open intervals like (0, 1), (0, 1/2), (0, 1/3), ... is still an open set (in this case, (0, 1)).

• Subsets of R²: Consider the open unit disk in R² (all points (x,y) such that x² + y² < 1). This can be represented as a union of infinitely many small open disks within the unit disk, demonstrating the theorem's application in higher dimensions.

This seemingly simple theorem forms the bedrock of many more advanced results in topology and analysis. Understanding its proof and implications is a crucial step in mastering these fields.