Question

Prove: (a) If \(U\) is a neighborhood of \(u_{0}\) and \(U \subset V,\) then \(V\) is a neighborhood of \(u_{0}\). (b) If \(U_{1}, U_{2}, \ldots, U_{n}\) are neighborhoods of \(u_{0}\), so is \(\cap_{i=1}^{n} U_{i}\).

Short answer

Question: Prove the following statements for neighborhoods: (a) If \(U\) is a neighborhood of \(u_{0}\) and \(U \subset V,\) then \(V\) is a neighborhood of \(u_{0}\). (b) If \(U_{1}, U_{2}, \ldots, U_{n}\) are neighborhoods of \(u_{0}\), then the intersection of these sets \(\cap_{i=1}^{n} U_{i}\) is also a neighborhood of \(u_{0}\). Answer: (a) For statement (a), we showed that when \(U\) is a neighborhood of \(u_{0}\), and \(U \subset V\), there exists an open set containing \(u_{0}\), \(O_{1} \subset U\), which is also a subset of V. This implies that \(V\) is a neighborhood of \(u_{0}\). (b) For statement (b), we showed that for a given set of neighborhoods \(U_{1}, U_{2}, \ldots, U_{n}\) of \(u_{0}\), the intersection of these sets, \(\cap_{i=1}^{n} U_{i}\), is also a neighborhood of \(u_{0}\) since there exists an open set containing \(u_{0}\), \(O\), which is a subset of the intersection.

Step-by-step solution

Prove statement (a)

Let \(U\) be a neighborhood of \(u_{0}\) and \(U \subset V\). By definition of a neighborhood, there exists an open set \(O_{1}\) containing \(u_{0}\) such that \(O_{1} \subset U\). Now, since \(U \subset V\), we have \(O_{1} \subset V\). Therefore, \(V\) is a neighborhood of \(u_{0}\) because we found an open set \(O_{1}\) containing \(u_{0}\) that is a subset of \(V\). Hence, we have proven statement (a).

Prove statement (b)

Let \(U_{1}, U_{2}, \ldots, U_{n}\) be neighborhoods of \(u_{0}\). By definition of a neighborhood, for each \(i\), there exists an open set \(O_{i}\) containing \(u_{0}\) such that \(O_{i} \subset U_{i}\). By construction, the set \(O= \cap_{i=1}^{n} O_{i}\) is an open set containing \(u_{0}\). Moreover, since \(O_{i} \subset U_{i}\), we have \(O \subset \cap_{i=1}^{n} U_{i}\). Therefore, \(\cap_{i=1}^{n} U_{i}\) is a neighborhood of \(u_{0}\) since we found an open set \(O\) containing \(u_{0}\) which is a subset of \(\cap_{i=1}^{n} U_{i}\). Hence, we have proven statement (b).

Key concepts

Understanding Neighborhoods in Real Analysis

In real analysis, the concept of a **neighborhood** of a point is quite important. A neighborhood of a point \( u_{0} \) in a set means you can find a little room or space around \( u_{0} \) within that set. This neighborhood includes an open interval (if we are on the real number line) or an open set (in general spaces) that covers \( u_{0} \).

Think of it as a circle drawn around a point on a map. The point is \( u_{0} \), and the circle gives you some space to move around without leaving the boundaries of what's considered the neighborhood. Mathematically, for a set \( U \) to be a neighborhood of \( u_{0} \), there must be an open set \( O \) such that \( u_{0} \in O \subset U \).

An interesting property is, if you have this little room and you decide to make it bigger or include it into a larger space, any larger space that includes this neighborhood will also be a neighborhood. This is essentially what statement (a) of the exercise was highlighting. If the smaller neighborhood \( U \) is part of a bigger set \( V \), anything true about \( U \) stays true for \( V \). It means \( V \) must also provide some bump room around \( u_{0} \). So, larger sets do not destroy the property of being a neighborhood.

Defining Open Sets and Their Relation to Neighborhoods

An **open set** is a crucial building block in real analysis. It's like an unbroken area or a space that doesn’t include its edge points, allowing complete liberty to move around within.

When an open set contains the point \( u_{0} \), it means you can move in any "small" amount from \( u_{0} \) and still be inside the set. Therefore, open sets provide the perfect building blocks for neighborhoods, because you need that kind of freedom around \( u_{0} \) to satisfy the definition of a neighborhood.

In our exercise, each neighborhood of \( u_{0} \) had an open set inside it. This open set guarantees that there is ample space around \( u_{0} \). When proving statement (b), even the intersection of several neighborhoods still retains \( u_{0} \). This intersection will also contain an open set reflecting the combined space from each neighborhood set \( U_{i} \).

So, open sets form an extensive network that interconnects neighborhoods, and this is what the solution is leveraging. Understanding open sets means knowing that wherever they nestle, they offer the freedom to wiggle around points like \( u_{0} \), ensuring the concept of a neighborhood stands strong.

Crafting a Proof in Real Analysis

A **proof** in real analysis is like a mathematical argument built through logic and steps. When proving the two parts of the exercise, we build on known truths and definitions to make valid conclusions.

For statement (a), the proof relies on understanding how neighborhoods function. Given \( U \) is a neighborhood of \( u_{0} \), and \( U \subset V \), it demands a specific order of logic:

• If there is an open set \( O_{1} \) that fits inside \( U \) and covers \( u_{0} \),
• Then \( O_{1} \), being open, will still be snug inside \( V \).

Thus, by showing there's an open set \( O_{1} \subset V \) containing \( u_{0} \), we establish that \( V \) inherits the neighborhood status.

For statement (b), where a collection of neighborhoods \( U_{1}, U_{2}, \, \ldots, U_{n} \) are united, the proof method is about intersection:

• If every \( U_{i} \) has an open set \( O_{i} \) that covers \( u_{0} \),
• The intersection of all these \( O_{i} \)'s remains open and covers \( u_{0} \).

Both proofs follow a path of logical connections: knowing what defines a neighborhood and how open sets ensure that a neighborhood's key features, like open space around \( u_{0} \), are given even when contexts change or combine.