Question

If \(A\) is a dense set in a topological space \(X\) and \(U \subseteq X\) is open, show that \(U \subseteq \overline{A \cap U}\).

Short answer

The given statement \(U \subseteq \overline{A \cap U}\) is proved using the definitions of dense set, open set, and closure. Our proof concludes that for every \(x \in U\), we have \(x \in \overline{A \cap U}\), which shows \(U \subseteq \overline{A \cap U}\).

Step-by-step solution

Analyzing \(U \subseteq \overline{A \cap U}\)

We aim to show that every element x of U is also an element of \(\overline{A \cap U}\), where \(\overline{A \cap U}\) denotes the closure of the set \(A \cap U\). The expression \(A \cap U\) represents intersection of set \(A\) and \(U\), so the points of \(A \cap U\) are the points which are common to both \(A\) and \(U\).

Utilizing the definition of a dense set

Since \(A\) is given as a dense set in \(X\), by its definition, the smallest closed set containing \(A\) is the entire topological space \(X\). Therefore, this includes the open set \(U\). Hence, the closure of set \(A\), \(\overline{A}\), contains \(U\), that is \(U \subseteq \overline{A}\).

Utilizing the definition of closure

The definition of the closure of a set includes all limit points of the set, i.e., for a point to be in the closure of a set, every open set containing the point must intersect the set. So the points in \(\overline{A \cap U}\) intersect \(A\) and \(U\), meaning that if \(x \in U\), then \(x \in \overline{A \cap U}\).

Concluding the proof

Bringing these steps together, we have shown that for every \(x \in U\), \(x \in \overline{A \cap U}\). Hence, \(U \subseteq \overline{A \cap U}\), which completes the proof as required.

Key concepts

Dense Sets

In the realm of topology, dense sets hold an intriguing role. A set \(A\) in a topological space \(X\) is termed dense if its closure equals the entire space. In simpler words, \(A\) is dense in \(X\) if you can find points of \(A\) really close to every point in \(X\).
This is akin to the idea that if you zoom in enough, no point in \(X\) is entirely isolated from \(A\). Thus, every neighborhood contains at least one point of \(A\).

• This brings us to understand that \(A\), when dense, means there aren't gaps between its elements in relation to \(X\).
• It overlaps all open sets in \(X\), essentially meaning \(U \cap \overline{A} eq \emptyset\) for any open set \(U\).

Closure of a Set

The closure of a set \(B\) in a topological space encompasses not only the points of \(B\) but also its limit points. Imagine the closure as the result of filling in any 'holes' that might leave \(B\) incomplete in terms of connectivity and proximity.
In notation, the closure is expressed as \(\overline{B}\). This concept is crucial because it formalizes the notion of 'border' points that can touch or extend \(B\).

• If a point lies in \(\overline{B}\), every neighborhood around that point hits \(B\) at some point.
• This implies that the closure of \(A \cap U\), \(\overline{A \cap U}\), is the smallest closed set that covers all of both \(A\) and \(U\).

Open Sets

Open sets are fundamental building blocks in topology. Put simply, an open set is one that does not include its boundary. Think of them as 'flexible' containers that can shape themselves freely without being tightly shut.
They are defined in such a way that for any point inside the set, there is a tiny bubble around it that fits entirely within the set. This provides a sense of local freedom and flexibility.

• An open set \(U\) in \(X\) means \(U\) is part of the topology on \(X\), making it officially recognized for use in analysis.
• Open sets allow mathematicians to talk about continuity, limits, and convergence with precision, setting the stage for deeper exploration into the interactions of points in \(X\).