Question

A set \(S\) is. in a set \(T\) if \(S \subset T \subset \bar{S}\). (a) Prove: If \(S\) and \(T\) are sets of real numbers and \(S \subset T,\) then \(S\) is dense in \(T\) if and only if every neighborhood of each point in \(T\) contains a point from \(S\). (b) State how (a) shows that the definition given here is consistent with the restricted definition of a dense subset of the reals given in Section \(1.1 .\)

Short answer

In summary, we have proved that if a set S is dense in set T, every neighborhood of each point in T contains a point from S, and vice versa. Additionally, we have shown that this definition is consistent with the definition of a dense subset of the real numbers given in Section 1.1, where a subset is dense if there exists a sequence of points in the subset converging to every real number.

Step-by-step solution

Proof (a) - Part 1: If \(S\) is dense in \(T\), then every neighborhood of each point in \(T\) contains a point from \(S\).

Since \(S\) is dense in \(T\), by definition, the closure of \(S\) (\(\bar{S}\)) includes \(T\). Therefore, if we take a point \(x \in T\), it must either be an element of \(S\) or a limit point of \(S\). In either case, for any open neighborhood \(N\) of \(x\), there exists a point \(y \in S\) such that \(y \in N\). Hence, every neighborhood of each point in \(T\) contains a point from \(S\).

Proof (a) - Part 2: If every neighborhood of each point in \(T\) contains a point from \(S\), then \(S\) is dense in \(T\).

Suppose every neighborhood of each point in \(T\) contains a point from \(S\). Now, we want to prove that the closure of \(S\) (\(\bar{S}\)) includes \(T\). Take any point \(x \in T\). Since every neighborhood of \(x\) contains a point from \(S\), \(x\) must be either an element of \(S\) or a limit point of \(S\). Thus, \(x \in \bar{S}\). Therefore, \(\bar{S}\) includes \(T\), which implies that \(S\) is dense in \(T\). Hence, we have proved part (a) of the exercise.

Statement (b): Relating part (a) to the restricted definition of a dense subset of the reals given in Section 1.1.

In Section 1.1, a subset \(S\) of the real numbers is defined to be dense if for every real number \(x\), there exists a sequence of points in \(S\) converging to \(x\). This definition states that, given any real number \(x\) and an open neighborhood \(N\) of \(x\), we can find a point in \(S\) that belongs to \(N\). This is because sequences convergence implies the existence of a sequence of points in \(S\) that get arbitrarily close to \(x\). Comparing this definition to part (a), we can see that they are consistent. Part (a) states that in order for \(S\) to be dense in \(T\), every neighborhood of each point in \(T\) must contain a point from \(S\). The restricted definition in Section 1.1 conveys the same idea, stating that a dense subset of the reals must have points in the neighborhood of every real number. Hence, the definition given in the exercise is consistent with the restricted definition of a dense subset of the reals given in Section 1.1.

Key concepts

Dense Subset

In the context of real analysis, a set \( S \) is said to be dense in another set \( T \) if every point in \( T \) is either a point of \( S \) or a limit point of \( S \). This essentially means that \( S \) overwhelmingly populates \( T \) at a granular level.

Let's break down the concept further: for \( S \) to be dense in \( T \), for any point \( x \) in \( T \), you can always find a point from \( S \) arbitrarily close to \( x \). This is akin to saying no matter how small a neighborhood you consider around any element of \( T \), it will always contain at least one point from \( S \).

When applied specifically to the set of real numbers, the set of rational numbers \( \mathbb{Q} \) serves as a classic example of a dense subset of the real numbers \( \mathbb{R} \), as between any two real numbers, no matter how close they are, a rational number can be found.

Closure of a Set

The closure of a set \( S \), denoted \( \bar{S} \), is an essential concept in topology and real analysis. It comprises all points in \( S \) together with all the limit points of \( S \). A limit point of \( S \) is a point that can be "approached" by other points of \( S \) but is not necessarily in \( S \) itself.

In practical terms, imagine drawing a boundary around the set \( S \). The closure \( \bar{S} \) would include every element of \( S \) as well as any point where you can approach from \( S \) but miss if you were to stop short.

This wraps \( S \) tightly with all possible limit points, effectively "closing" it. Understanding closure helps in visualizing how \( S \) might fit or nest within a larger set \( T \). If \( T \subset \bar{S} \), it means \( S \) can be considered dense in \( T \), because every element of \( T \) becomes an element of the closure of \( S \).

Neighborhood in Topology

In topology, a neighborhood of a point \( x \) isn't limited to containing \( x \) itself but instead is any set that contains an open set including \( x \). Visualize a neighborhood as a small "bubble" around a point that holds everything inside it closer than any point outside.

Open sets, in this respect, are collections of points that do not include their boundary, allowing neighborhoods to extend various sizes and shapes as long as they meet this condition.

Topologically speaking, neighborhoods assist in explaining concepts like continuity and limits by grounding them in spatial terms. They are key in the definition of dense subsets because, to be dense, a set \( S \) must intersect every neighborhood of each point in the set it densifies \( T \).

• A neighborhood of a point \( x \) will usually be an epsilon "distance" around \( x \)
• The real-valued intervals \( (x - \epsilon, x + \epsilon) \) serve as straightforward examples of neighborhoods for points on the real line.
• This detailed notion helps affirm when and how clusters of points form around specific points, essential in dense subsets and closure.