Question

Show \(X\) is a separable normed linear space if and only for each \(\epsilon>0\) there exists a countable collection of sets of diameter at most \(\epsilon\) such that \(X\) is continued in their union.

Short answer

In summary, a normed linear space \(X\) is separable if and only if for each \(\epsilon > 0\), there exists a countable collection of sets with diameter at most \(\epsilon\) such that \(X\) is contained in their union. We proved this by first assuming separability and showing the existence of the countable collection using a countable dense subset and open balls. Then, we assumed the existence of the countable collection and showed that the space is separable by constructing a countable dense set from the collection.

Step-by-step solution

Part 1: Assume separability and show the existence of the collection.

Let \(X\) be a separable normed linear space. By definition, it contains a countable dense subset, say \(D = \{x_n\}_{n=1}^{\infty}\). Now, let \(\epsilon > 0\). For each \(n \in \mathbb{N}\), consider the open ball with diameter \(B(x_n, \frac{\epsilon}{2}) = \{x \in X: \|x - x_n\| < \frac{\epsilon}{2}\}\). The union of these balls, \(\bigcup_{n=1}^{\infty} B(x_n, \frac{\epsilon}{2})\), is a countable collection of sets that covers \(X\), since \(D\) is dense in \(X\). For any two points \(x, y \in B(x_n, \frac{\epsilon}{2})\), by the triangle inequality, we have \(\|x - y\| \le \|x - x_n\| + \|x_n - y\| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\). Therefore, each set \(B(x_n, \frac{\epsilon}{2})\) has diameter at most \(\epsilon\).

Part 2: Assume the existence of the collection and show that X is separable.

Now, let's suppose that for every \(\epsilon > 0\), there exists a countable collection of sets of diameter at most \(\epsilon\) such that \(X \subseteq \bigcup_{n=1}^{\infty} A_n\). Fix \(\epsilon > 0\). Choose one point from each set \(A_n\), say \(x_n\). Let \(D = \{x_n\}_{n=1}^{\infty}\). Then, \(D\) is a countable set. We will now show that \(D\) is dense in \(X\). Let \(x \in X\) and \(\delta > 0\) be arbitrary. Since \(X \subseteq \bigcup_{n=1}^{\infty} A_n\) and each \(A_n\) has diameter at most \(\frac{\delta}{2}\), there exists an \(n \in \mathbb{N}\) such that \(x \in A_n\). By our choice of \(D\), we have that \(x_n \in A_n\). Therefore, \(\|x - x_n\| \le \mathrm{diam}(A_n) \le \frac{\delta}{2} < \delta\). This shows that \(D\) is dense in \(X\). Therefore, \(X\) is a separable normed linear space.

Key concepts

Countable Dense Subset

Understanding the concept of a countable dense subset is crucial in the context of separable normed linear spaces. A set is said to be dense in a space if every point in that space is either in the dense set or is a limit point of the dense set. That means any point in the space can be approximated arbitrarily closely by points from the dense set.

Now, when we say a dense set is countable, it means that it has the same size as the natural numbers, which implies it is either finite or can be put into a one-to-one association with the set of natural numbers.

In a separable normed linear space, having a countable dense subset helps simplify complex structures by allowing us to approximate any point in the space by points from this simpler and more manageable countable set.

Open Balls in Metric Spaces

Open balls are fundamental building blocks in understanding the structure of metric spaces. An open ball centered at a point within a space, with a certain radius, is a set of all points that are closer to the center than the given radius.

Mathematically, for a point \(x_n\) in a space and a radius \(r\), an open ball \(B(x_n, r)\) is defined as \( \{ x \in X: \|x - x_n\| < r \} \). This means that all elements \(x\) in the open ball are within a distance of \(r\) from the center \(x_n\).

In our exercise, using open balls with a diameter \(\frac{\epsilon}{2}\) helps in systematically covering the entire space \(X\) with countable collections of these smaller, manageable sets. Each open ball acts as one piece that collectively, with others, "touches" every point in \(X\), ensuring that no point remains uncovered.

Triangle Inequality

The triangle inequality is a simple yet powerful tool in mathematics, especially in the context of metric spaces. It states that, for any three points, the direct path between two points is always shorter or equal to any indirect path visiting a third point.

In mathematical terms, for points \(x\), \(y\), and \(z\), the triangle inequality is expressed as \(\|x - y\| \leq \|x - z\| + \|z - y\|\).

In the context of open balls and showing their coverage, the triangle inequality assists us by proving that any two points in the same open ball cannot be further apart than the maximum diameter of that ball. Consequently, if a ball has a radius of \(\frac{\epsilon}{2}\), any two points \(x\) and \(y\) inside satisfying \(\|x - y\| < \epsilon\), helping us to maintain the defined conditions of separability in metric spaces.

Dense Sets

Dense sets are sets whose closure is the entire space in which they reside. This means they come arbitrarily close to every point in the space, effectively allowing no gaps.

A dense set \(D\) in space \(X\) ensures any point \(x\) in \(X\) can be approached as closely as desired by points in \(D\). If \(D\) is countable and dense, it not only covers the space with overlapping circles but does so in a way that maintains a clear structure (a sequence) due to its countable nature.

For a set \(D\) to be considered dense, you can take any small positive number (no matter how small) and still find that points from \(D\) are within that distance from any point in \(X\). Such a setup is foundational to understanding the separability of spaces, as a separable space inherently has this countable dense characteristic.