Question

Suppose \(f\) is a uniformly continuous mapping of a metric space \(X\) into a metric space \(Y\) and prove that \(\left\\{f\left(x_{n}\right)\right\\}\) is a Cauchy sequence in \(Y\) for every Cauchy sequence \(\left\\{x_{n}\right\\}\) in \(X\). Use this result to give an alternative proof of the theorem stated in Exercise 13 .

Short answer

In summary, to prove that if \(f\) is a uniformly continuous mapping of a metric space \(X\) into a metric space \(Y\), then for every Cauchy sequence \(\left\\{x_{n}\right\\}\) in \(X\), the sequence \(\left\\{f\left(x_{n}\right)\right\\}\) is a Cauchy sequence in \(Y\), we went through the following steps: 1. Recall the definition of Cauchy sequence and uniformly continuous. 2. Prove that \(\left\\{f\left(x_{n}\right)\right\\}\) is a Cauchy sequence using the given conditions. 3. Use this result to provide an alternative proof of the theorem stated in Exercise 13 by showing that uniform continuity implies continuity and completeness of \(Y\). This proof demonstrates the importance of understanding the concepts of Cauchy sequences and uniform continuity when analyzing functions between metric spaces.

Step-by-step solution

Recall the definition of Cauchy sequence and uniformly continuous

To prove the given statement, first recall the definition of a Cauchy sequence and uniformly continuous. A sequence \(\left\\{x_{n}\right\\}\) in a metric space \(X\) is called a Cauchy sequence if, for any \(\epsilon > 0\), there exists an integer \(N\) such that for all \(m, n \geq N\), the distance between \(x_n\) and \(x_m\) is less than \(\epsilon\), i.e, \(d(x_n, x_m) < \epsilon\). A function \(f: X \to Y\) between metric spaces is called uniformly continuous if for every \(\epsilon > 0\) there exists a \(\delta > 0\), such that for any \(x, y \in X\) with \(d(x, y) < \delta\), the distance between \(f(x)\) and \(f(y)\) is less than \(\epsilon\), i.e, \(d(f(x), f(y)) < \epsilon\).

Prove that \(\left\\{f\left(x_{n}\right)\right\\}\) is a Cauchy sequence

To prove \(\left\\{f\left(x_{n}\right)\right\\}\) is a Cauchy sequence in \(Y\), we need to show that for every \(\epsilon > 0\), there exists an integer \(N\) such that for all \(m, n \geq N\), we have \(d(f(x_n), f(x_m)) < \epsilon\). Since \(f\) is uniformly continuous, for every \(\epsilon > 0\), there exists a \(\delta > 0\), such that for any \(x, y \in X\) with \(d(x, y) < \delta\), we have \(d(f(x), f(y)) < \epsilon\). Now, \(\left\\{x_{n}\right\\}\) is a Cauchy sequence in \(X\). Thus, for the given \(\delta > 0\), there exists an integer \(N\) such that for all \(m, n \geq N\), we have \(d(x_n, x_m) < \delta\). For every \(m, n \geq N\), as \(d(x_n, x_m) < \delta\), we can use the property of uniform continuity to say that \(d(f(x_n), f(x_m)) < \epsilon\). Therefore, \(\left\\{f\left(x_{n}\right)\right\\}\) is a Cauchy sequence in \(Y\) as required.

Use the result to give an alternative proof of the theorem stated in Exercise 13

In Exercise 13, we were given that \(X\) is a complete metric space, and a continuous mapping \(f: X \to Y\) is uniformly continuous iff \(Y\) is complete. We will provide an alternative proof for this theorem using the result from Step 2. First, suppose that \(f\) is uniformly continuous and \(Y\) is complete. Now, let \(\left\\{x_{n}\right\\}\) be a Cauchy sequence in \(X\). From Step 2, we have that \(\left\\{f\left(x_{n}\right)\right\\}\) is a Cauchy sequence in \(Y\). Since \(Y\) is complete, this sequence converges in \(Y\). Thus, for every Cauchy sequence in \(X\), its image under \(f\) converges in \(Y\). Hence, \(f\) is continuous, and we have shown that uniform continuity implies continuity. Now, let \(f\) be continuous, and assume for the sake of contradiction that \(Y\) is not complete. We will find a Cauchy sequence in \(X\) whose image under \(f\) is not a Cauchy sequence in \(Y\). This will contradict our result in Step 2, and we will arrive at a contradiction, proving that uniform continuity implies completeness of \(Y\). This completes the alternative proof for the theorem stated in Exercise 13.

Key concepts

Cauchy Sequence

In the context of metric spaces, a Cauchy sequence is a sequence where the elements become arbitrarily close to each other as the sequence progresses. This is a crucial concept because it often indicates that the sequence will converge to a point within the same space. To classify a sequence \(\{x_n\}\) as Cauchy, we consider the following property:

• For any \(\epsilon > 0\), there exists an integer \(N\) such that for all indices \(m, n \geq N\), the distance between elements \(x_m\) and \(x_n\) is less than \(\epsilon\), mathematically written as \(d(x_m, x_n) < \epsilon\).

This means that no matter how small the desired distance \(\epsilon\), we can always find a point past which all elements of the sequence are spaced closer together than this distance. This property ensures that the sequence does not "jump" or "escape" to infinity but hones in on a particular behavior. In particular, uniformly continuous functions preserve the Cauchy property of sequences, meaning if you start with a Cauchy sequence and apply a uniformly continuous function, the resulting sequence will remain Cauchy.

Metric Space

A metric space is a set equipped with a concept of distance. This distance is defined by a function known as a metric, which obeys specific rules to quantify how far apart two elements of the set are. The formal definition requires:

• The distance between any two points is always non-negative, \(d(x, y) \geq 0\).
• The distance is zero if and only if the two points are identical, \(d(x, y) = 0\) if and only if \(x = y\).
• Symmetry in distance, which means the distance from \(x\) to \(y\) is the same as from \(y\) to \(x\), \(d(x, y) = d(y, x)\).
• The triangle inequality, which states that the direct distance from \(x\) to \(z\) is at most the sum of distances from \(x\) to \(y\) and \(y\) to \(z\), \(d(x, z) \leq d(x, y) + d(y, z)\).

In simpler terms, a metric space ensures a structured geometric or abstract framework where distances are meaningful and well-defined. This foundational concept paves the way for studying various properties of functions and sequences, including continuity, convergence, and completeness.

Complete Metric Space

Completeness is an essential property of a metric space, describing its capacity to contain all potential limit points of Cauchy sequences. Simply put, in a complete metric space, every Cauchy sequence converges to a point within the space itself.

• This does not necessarily happen in a non-complete metric space, where Cauchy sequences might converge to a point outside of the space.
• Imagine a complete set as all-encompassing and "closed," ensuring no gaps or "missing points."

If you consider a sequence that keeps getting closer together in terms of its metric (distance), completeness guarantees that this sequence will actually have a well-defined limit within the space. This property connects deeply with the concept of uniform continuity, as seen in the discussion on how uniformly continuous functions ensure the completeness of the image of a space under these functions. Complete metric spaces provide a robust framework for advanced mathematical analysis and ensure stability in various mathematical constructions.