91Ó°ÊÓ

Limit points are a foundational concept in topology and analysis. A limit point of a set is a point that can be "approached" by other points of the set. Formally, a point \( x \) is a limit point of a set \( M \) if every neighborhood of \( x \) contains at least one point of \( M \) different from \( x \) itself. For example, if you have a sequence that gets closer and closer to a number, that number is a limit point.

The significance of limit points in this exercise is that they help form the topological closure of a set. The closure, denoted as \( \bar{M} \), includes all limit points of all convergent sequences from the set \( M \). Thus, \( \bar{M} \) essentially "completes" \( M \) by including its boundary or edge points where sequences in \( M \) converge.

A sequence is a list of numbers arranged in a particular order. A convergent sequence is one where the terms get closer and closer to a specific value, known as the limit, as the sequence progresses. In mathematical terms, a sequence \( (a_n) \) of elements in a set \( M \) converges to \( L \) if for every positive number \( \epsilon \), there exists an integer \( N \) such that for all \( n > N \), \(|a_n - L| < \epsilon\). This means the terms of the sequence can be made arbitrarily close to \( L \) by making \( n \) large enough.

In the context of the exercise, the set \( \bar{M} \) contains all limit points of convergent sequences from \( M \). This means that any point which can be reached as a limit of a sequence of points in \( M \) is part of \( \bar{M} \). This property allows \( \bar{M} \) to capture all possible "points of approach" originating from \( M \).

A closed set is a type of set in mathematics that includes all its limit points. No point "near" the set is excluded; it encompasses boundaries, making it distinct from open sets which do not include boundary points. A set \( A \) is closed if every limit point of \( A \) is contained in \( A \) itself.

In this exercise, proving that \( \bar{M} \) is closed involves showing it contains all its limit points. By definition, since \( \bar{M} \) includes all limits of sequences from \( M \), any further limits derived from sequences in \( \bar{M} \) must also belong to \( \bar{M} \). Thus, \( \bar{M} \) can't "omit" any boundary point, making it inherently closed.

Subset inclusion is a fundamental concept in set theory, where one set is said to be a subset of another if all the elements of the first set are contained in the second set. Symbolically, \( A \subseteq B \) implies every element of \( A \) is also an element of \( B \). This inclusion is significant in the context of this exercise.

Demonstrating that \( M \subseteq \bar{M} \) means showing that every point in \( M \) can naturally be viewed as a limit point in \( \bar{M} \). This is straightforward since any element \( x \in M \) forms a trivial constant sequence that converges to \( x \) itself. Therefore, it must lie within \( \bar{M} \). Consequently, proving the inclusion provides that the closure \( \bar{M} \) contains the original set \( M \) as part of its completion.