Question

If \(X\) is normal and second countable, there is a countable family \(\mathcal{F} \subset C(X, I)\) that separates points and closed sets. (Let \(\mathcal{B}\) be a countable base for the topology. Consider the set of pairs \((U, V) \in \mathcal{B} \times \mathcal{B}\) such that \(\bar{U} \subset V\), and use Urysohn's lemma.)

Short answer

A countable family \(\mathcal{F}\) is constructed using Urysohn's functions to separate points and closed sets.

Step-by-step solution

Understanding the Problem

We are asked to demonstrate that there exists a countable family \(\mathcal{F} \subset C(X, I)\) that separates points and closed sets in a space \(X\), which is both normal and second countable. \(C(X, I)\) denotes the set of continuous functions from \(X\) to the closed interval \([0, 1]\), and we can use properties of normal and second countable spaces, as well as Urysohn's lemma.

Identify Countable Base

Since \(X\) is second countable, there exists a countable base \(\mathcal{B}\) for the topology of \(X\). This means that every open set in \(X\) can be expressed as a union of elements from this countable collection \(\mathcal{B}\).

Define Urysohn Pair Set

Consider the set of pairs \((U, V) \in \mathcal{B} \times \mathcal{B}\) such that \(\bar{U} \subset V\), where \(\bar{U}\) denotes the closure of \(U\). This condition ensures \(U\) and \(V\) can be utilized with Urysohn's lemma because \(X\) is normal. The set of all such pairs is countable because it is a subset of the product \(\mathcal{B} \times \mathcal{B}\), which is countable.

Apply Urysohn's Lemma

For each pair \((U, V)\), Urysohn's lemma provides a continuous function \(f_{UV} : X \rightarrow [0, 1]\) such that \(f_{UV}(x) = 0\) for \(x \in \bar{U}\) and \(f_{UV}(x) = 1\) for \(x \in X \setminus V\). These functions are specifically chosen to "separate" the closed set \(\bar{U}\) and the open set \(X \setminus V\).

Create Countable Family

Define \(\mathcal{F} = \{ f_{UV} \mid (U,V) \in \mathcal{B} \times \mathcal{B}, \bar{U} \subset V \}\). This family is countable because it corresponds to the countable set of pairs \((U, V)\). Since \(f_{UV}\) are continuous and separate points and closed sets, \(\mathcal{F}\) satisfies the needed properties.

Conclusion

Thus, we have constructed a countable family \(\mathcal{F} \subset C(X, I)\) that separates points and closed sets in \(X\), utilizing the properties of second countable and normal spaces along with Urysohn's lemma.

Key concepts

Second Countable Space

Second countable spaces are a special type of topological space that come with a neat and compact characteristic: they have a **countable base**. This simply means that there exists a countable collection of open sets such that any open set in the space can be formed as a union of these sets.
Second countability is a powerful property because it simplifies many problems by reducing them to a countable set, which is often easier to manage and understand.
Some of the benefits of second countability include:

• Any second countable space is separable, meaning it contains a countable, dense subset.
• Second countability ensures that the space has a **metric**, making it more manageable for analysis and topology.
• It plays a role in ensuring that continuous functions behave well, especially when paired with other properties like normality.

In the given exercise, the second countability of space \(X\) allows us to use a countable base \(\mathcal{B}\) which proves crucial for constructing the countable family \(\mathcal{F}\) of continuous functions that helps in separating points and closed sets.

Urysohn's Lemma

Urysohn's Lemma is a beautiful and practical tool in topology that applies to normal spaces. A normal space is one where we can "separate" any two disjoint closed sets with disjoint open neighborhoods. Urysohn's Lemma takes this concept a step further.
It states that if you have two disjoint closed sets \(A\) and \(B\) within a normal space, there exists a continuous function \(f : X \rightarrow [0, 1]\) such that \(f(x) = 0\) for every point \(x\) in set \(A\) and \(f(x) = 1\) for every point \(x\) in set \(B\). The function smoothly transitions between \(0\) and \(1\) over \(X\), demonstrating the ability to continuously separate the two sets.
In our context, applying Urysohn's Lemma allows us to construct specific continuous functions \(f_{UV}\) for pairs of sets \((U, V)\), enabling us to separate closed subsets of \(X\) using this elegant approach. This is key to forming the set \(\mathcal{F}\) of functions in the solution, linking the idea to the countable base \(\mathcal{B}\).

Continuous Functions

Continuous functions are at the heart of topology, helping to relate spaces in a meaningful way. For a function \(f : X \rightarrow Y\) to be continuous, every time you take a point \(x\) and a sequence of points approaching \(x\), the images of these points under \(f\) should also approach the image \(f(x)\). This idea ensures there's no "jump" in the function.
In topology, continuous functions have profound implications, including the ability to transport topological properties from one space to another and perform algebraic operations that preserve topology.
Here are some key points about continuous functions:

• They map open sets to open sets, a property that maintains the structure of the original space.
• Their smoothness makes them perfect for transitioning between spaces, an essential aspect in applied mathematics and physics.
• They aid in the creation and understanding of complex spaces through simple transformations.

In this exercise, continuous functions \(f_{UV}\), defined via Urysohn's Lemma, handle the separation of closed sets and the exploration of the topology within the normal and second countable space \(X\). These functions aren't just abstract tools, but practical means to measure and interact with the space's properties efficiently.