Problem 9 - An Introduction to Abstract Analysis

Question

Let \(X\) be complete, and \(\left\\{F_{n}\right\\}\) a sequence of continuous functions from \(X\) to \(\mathbb{R}\) with the property that \(f_{n}(x) \rightarrow f(x)\) as \(n \rightarrow \infty\) for all \(x \in X\). Prove that the set of points in \(X\) at which \(f\) is continuous is dense in \(X\). [Hint: let $$ A_{m n}=\left\\{x:\left|f_{p}(x)-f_{n}(x)\right| \leq 1 / m \text { for all } p \geq n\right\\} $$ Show using the previous question that \(G_{m}=\cup_{n} \operatorname{int} A_{m n}\) is dense in \(X\), and prove that \(f\) is continuous at each point of \(\cap_{m} G_{m}\).]

Step-by-Step Solution

Answer

In conclusion, given a complete metric space \(X\) and a sequence of continuous functions \(\left\\{F_{n}\right\\}\) from \(X\) to \(\mathbb{R}\) such that \(f_n(x) \rightarrow f(x)\), we have shown that the set of points in \(X\) where \(f\) is continuous is dense in \(X\). We achieved this conclusion by defining sets \(A_{mn}\) and \(G_m\), showing that \(G_m\) is dense in \(X\), and proving that \(f\) is continuous at each point of \(\cap_{m} G_{m}\).

Step 1: Define the sets \(A_{mn}\) and \(G_m\).

Based on the hint, we define the sets as follows: $$ A_{mn} = \left\\{x \in X : |f_p(x) - f_n(x)| \leq \frac{1}{m} \text{ for all } p \geq n\right\\} $$ and $$ G_m = \bigcup_{n} \operatorname{int} A_{mn}. $$ Our goal is to show that \(G_m\) is dense in \(X\) and that \(f\) is continuous at each point of \(\cap_{m} G_{m}\).

Step 2: Prove that \(G_m\) is dense in \(X\).

By the Baire Category theorem, a countable union of nowhere dense sets cannot be dense. Therefore, to prove that \(G_m\) is dense in \(X\), we should show that each set \(\operatorname{int} A_{mn}\) is not a nowhere dense set, or equivalently, that \(A_{mn}\) has non-empty interiors for all \(m, n\). Now let's consider the continuity of the functions \(f_n\): Since \(f_n\) are continuous on \(X\), for each \(n\) and for every \(x \in X\) and for every \(\epsilon > 0\), there exists a neighborhood \(V\) of \(x\) such that: $$ |f_n(y) - f_n(x)| < \epsilon \quad \text{for all } y \in V. $$ As \(f_p(x) - f_n(x) \rightarrow 0\) for every \(x \in X\), we can make it smaller than any \(\epsilon\) by choosing a large enough \(p\). Thus, we can find some \(p \geq n\) such that \(|f_p(x) - f_n(x)| < \epsilon\) for any \(x \in X\). Now, let's set \(\epsilon = \frac{1}{m}\), so that we can choose neighborhoods \(V\) for each \(x\) that satisfy the continuity of \(f_n\), and hence show that each set \(A_{mn}\) has non-empty interiors. This implies that each \(\operatorname{int}A_{mn}\) is not nowhere dense. Therefore, \(G_m = \bigcup_{n} \operatorname{int} A_{mn}\) is dense in \(X\).

Step 3: Prove that \(f\) is continuous at each point of \(\cap_{m} G_{m}\).

Let's take a point \(x \in \cap_{m} G_{m}\). Then, \(x \in G_m\) for all \(m\). Therefore, there exists a sequence \(x_n \rightarrow x\) such that \(x_n \in \operatorname{int}A_{mn}\) for all \(n\). This means that $$ |f_p(x_n) - f_n(x_n)| \leq \frac{1}{m} \quad \text{for all } p \geq n. $$ As \(x_n \rightarrow x\), we have: $$ |f_p(x) - f_n(x)| \leq \frac{1}{m} + |f_p(x_n) - f_p(x)| + |f_n(x_n) - f_n(x)| \leq \frac{1}{m} + \epsilon. $$ Now, let \(p \rightarrow \infty\) and \(m \rightarrow \infty\). Then observe that \(|f_p(x) - f_n(x)| \rightarrow 0\). Thus, we can conclude that \(f\) is continuous at each point in \(\cap_{m} G_{m}\). Since \(G_m\) is dense in \(X\) and \(f\) is continuous at each point in \(\cap_{m} G_{m}\), we have that the set of points in \(X\) at which \(f\) is continuous is dense in \(X\).

Key Concepts

Understanding Pointwise Convergence and the Baire Category Theorem

Pointwise convergence is a concept in analysis where a sequence of functions \(f_n(x)\) converges to a function \(f(x)\) for each point \(x\) individually, as \(n \rightarrow \infty\). This is distinct from uniform convergence, where the convergence rate is the same across the entire domain. To delve deeper, consider a scenario where we have an assembly of continuous function sequences transitioning towards a limit function \(|f_{n}(x) \rightarrow f(x)|\) within a complete metric space \(X\).

The Baire Category Theorem is a powerful tool that aids in understanding the implications of pointwise convergence. This theorem, applicable to complete metric spaces, states that the union of countably many nowhere dense subsets cannot constitute the entire space. In simpler terms, 'large' parts of the space cannot be made up from 'small', 'insignificant' parts. When applied to our exercise, we leverage this theorem to illustrate that the set \(G_m\), defined as the union of the interiors of \(A_{mn}\), must be dense if \(A_{mn}\) is not nowhere dense, which in essence, arises from the continuity of the sequence of functions \(f_n\). By showing \(G_m\) is dense in \(X\), we deduce that so too is the set of points where \(f\) is continuous, a profound result growing directly from the Baire Category Theorem's implications.

Grasping Complete Metric Spaces

The backdrop of our problem relies on a setting within a complete metric space. But what makes a space 'complete'? A metric space is defined by a set of points along with a metric—a way to measure distances between these points. A space is complete when every Cauchy sequence, a sequence where elements get arbitrarily close to each other as the sequence progresses, has a limit that lies within the space itself.

In everyday terms, this means there are no 'holes' or missing points; imagine completing a puzzle only for it to include every single piece. The relevance of completeness arises when examining convergence properties within the space. For our example, the completeness of \(X\) ensures that the pointwise limit function \(f\) exists for each point in \(X\). Thus, guaranteeing that our analysis on convergence and continuity remains on solid footing, firmly within the confines of the space.

The Role of Continuous Functions

Now, let's focus on continuous functions, a key player in our exercise. A function \(f\) is continuous at a point \(x\) if small nudges to \(x\) only cause small ripples in the value of \(f(x)\). In mathematical language, for any degree of closeness (denoted by \(\epsilon\)) we desire in the function's output, we can find a closeness (denoted by \(\delta\)) in the input \(x\) to match it.

Why is this important? Because the very nature of continuous functions allows us to control their behavior on a small scale, which cascades to our ability to speak about their behavior on a larger scale. When the functions \(f_n\) in our sequence are continuous, we're equipped to prove that sets like \(A_{mn}\) possess non-empty interiors, thus enabling us to demonstrate that the points of continuity of the limit function \(f\) are abundant throughout the space, creating a dense network of continuity. It's like stitching a fabric with the thread of continuity; the more widespread the stitches, the more intact the fabric.