Question

The set \(\\{x\\}\), which consists of a single point in the normed linear space \(X\), is closed.

Short answer

Question: Prove that a set with a single point, \(\\{x\\}\), in a normed linear space \(X\) is closed. Answer: Since every open ball around the point \(x\) contains at least one point from the set different from \(x\), we can conclude that the point \(x\) is a limit point of the set \(\\{x\\}\). As the only limit point of the set is contained within the set itself, the set is closed in the normed linear space.

Step-by-step solution

Remind the definition of a limit point of a set in a normed linear space

In a normed linear space \(X\), a point \(y \in X\) is a limit point of a set \(A \subseteq X\) if every open ball around \(y\) contains at least one point of \(A\) different from \(y\).

Define an open ball centered at the point x

Now, let's consider an arbitrary open ball \(B(x;r)\) around the point \(x\), with radius \(r > 0\), where \(x\) is the only point in the set \(\\{x\\}\).

Show that every open ball around x contains a point from the set different from x

Since \(x \in \\{x\\}\) and the open ball \(B(x;r)\) is centered at \(x\), there exists at least one point of the set in the open ball different from \(x\). Now, we have to show that the open ball around the point x does not contain any other point of the set besides point x. Actually, there is no other point in the set \(\\{x\\}\). So, every open ball around the point x contains only the point x from the given set.

Conclude that the point x is a limit point of the set

Since every open ball around the point \(x\) contains at least one point from the set different from \(x\), we can conclude that the point \(x\) is a limit point of the set \(\\{x\\}\).

Conclude that the set is closed

As we have shown that the only limit point of the set \(\\{x\\}\) is contained within the set itself, we can conclude that the set is closed in the normed linear space.

Key concepts

Closed Set

A closed set in a normed linear space is an important concept in functional analysis. Understanding whether a set is closed helps in understanding the structure of the space.
In intuitive terms, a set is closed if it contains all its limit points.

• If no limit point exists outside the set, then the set is automatically closed.
• Mathematically, if a point is a limit point of the set, it must belong to the set for it to be closed.

In the exercise given, the set \(\{x\}\) is indeed closed, because it contains its only limit point, itself. This shows that even a single-point set is closed in a normed linear space.

Limit Point

The concept of a limit point is central to understanding sets in topological spaces, including normed linear spaces.
A point \(y\) is a limit point of a set \(A\) if every open ball around \(y\) contains at least one point from \(A\) different from \(y\) itself.

• This means there's always a sequence of points inside \(A\) that comes arbitrarily close to \(y\).
• This characterizes the set's boundary and helps determine whether it is closed or not.

In the given example, \(x\) in the set \(\{x\}\) is a limit point because there's no other point in the open ball except itself. All open balls centered at \(x\) will include \(x\), supporting its self-containment as a limit point.

Open Ball

An open ball is a fundamental geometric shape in metric and normed spaces, playing a pivotal role in analysis.
An open ball \(B(x;r)\) centered at a point \(x\) with radius \(r\) is the set of points which are less than \(r\) distance away from \(x\).

• The notion of open balls helps define both continuity and limit points within these spaces.
• They're used in formulating definitions and properties involving open and closed sets.

In the context of this problem, even though the set only contains \(x\), the open ball \(B(x;r)\) exhibits how all points within \(r\) distance are encompassed. Since \(x\) is the only point, the open ball always includes \(x\) alone, reinforcing \(x\)'s status as its own limit point and verifying the closed nature of the set \(\{x\}\).