Question

Suppose that \((A, \rho),(B, \sigma),\) and \((C, \gamma)\) are metric spaces, and let $$ f:(A, \rho) \rightarrow(B, \sigma) \quad \text { and } \quad g:(B, \sigma) \rightarrow(C, \gamma), $$ where \(D_{f}=A, R_{f}=D_{g}=B,\) and \(f\) and \(g\) are continuous. Define \(h:(A, \rho) \rightarrow\) \((C, \gamma)\) by \(h(u)=g(f(u)) .\) Show that \(h\) is continuous on \(A\).

Short answer

Question: Prove that if we define a new function \(h:(A, \rho) \rightarrow (C, \gamma)\) by the composition of two continuous functions \(f:(A, \rho) \rightarrow (B, \sigma)\) and \(g:(B, \sigma) \rightarrow (C, \gamma)\), then \(h\) is continuous on metric space \((A, \rho)\). Answer: To prove the continuity of \(h\), we need to show that for every point \(u \in A\) and for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(\rho(u, v) < \delta\), then \(\gamma(h(u), h(v)) < \epsilon\). By applying the continuity of \(f\) and \(g\), we can show that this condition holds for \(h\) at every point in \(A\), proving that \(h\) is continuous on \((A, \rho)\).

Step-by-step solution

Recall the definition of continuity in metric spaces

In order for a function to be continuous in metric spaces, we need to show that for every point \(u \in A\) and for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(\rho(u, v) < \delta\), then \(\gamma(h(u), h(v)) < \epsilon\).

Fix a point \(u \in A\) and an arbitrary \(\epsilon > 0\)

We will now take an arbitrary point \(u \in A\) and an arbitrary value \(\epsilon > 0\) as stated in the definition of continuity. We need to find a \(\delta > 0\) for this \(u\) and \(\epsilon\) that satisfies the continuity condition.

Apply continuity of \(g\) at \(f(u)\)

Since \(g\) is continuous in \((B, \sigma)\), then for point \(f(u) \in B\) and for every \(\epsilon > 0\), there exists a \(\epsilon_1 > 0\) such that if \(\sigma(f(u), y) < \epsilon_1\), then \(\gamma(g(f(u)), g(y)) < \epsilon\).

Apply continuity of \(f\) at \(u\)

Now, we know that \(f\) is continuous in \((A, \rho)\). Therefore, for the point \(u \in A\) and for every \(\epsilon_1 > 0\), there exists a \(\delta > 0\) such that if \(\rho(u, v) < \delta\), then \(\sigma(f(u), f(v)) < \epsilon_1\).

Show that the continuity condition holds for \(h\)

If we choose \(y = f(v)\), then we have: 1. If \(\rho(u, v) < \delta\), then \(\sigma(f(u), f(v)) < \epsilon_1\) (from Step 4). 2. If \(\sigma(f(u), f(v)) < \epsilon_1\), then \(\gamma(g(f(u)), g(f(v))) < \epsilon\) (from Step 3). Thus, if \(\rho(u, v) < \delta\), we have: $$\gamma(h(u), h(v)) = \gamma(g(f(u)), g(f(v))) < \epsilon$$ which is the continuity condition for \(h\) at point \(u\) in the metric space \((A, \rho)\).

Conclude that \(h\) is continuous

Since we have shown that for every point \(u \in A\) and for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(\rho(u, v) < \delta\), then \(\gamma(h(u), h(v)) < \epsilon\), we can conclude that the function \(h\) is continuous on metric space \((A, \rho)\).

Key concepts

Metric Spaces

A metric space is a fundamental concept in mathematics, specifically in topology and analysis. It consists of a set along with a metric, which is a function that defines a distance between any two elements in the set. In simpler terms, if we think of the set as a collection of points, the metric tells us how far apart these points are from each other.

Formally, for a set 'A', a metric is a function \(\rho : A \times A \rightarrow \mathbb{R}\) satisfying four key properties:

• Non-negativity: \(\rho(a, b) \geq 0\) for all \(a, b \in A\), and \(\rho(a, b) = 0\) if and only if \(a = b\).
• Symmetry: \(\rho(a, b) = \rho(b, a)\) for all \(a, b \in A\).
• The triangle inequality: \(\rho(a, c) \leq \rho(a, b) + \rho(b, c)\) for any \(a, b, c \in A\).
• Definiteness or separation: \(\rho(a, b) > 0\) if \(a eq b\).

In the exercise, \(f:(A, \rho) \rightarrow (B, \sigma)\) and \(g:(B, \sigma) \rightarrow (C, \gamma)\) indicate that \(A\), \(B\), and \(C\) are sets with their respective metrics \(\rho\), \(\sigma\), and \(\gamma\), making each a metric space.

Continuous Function

The term ‘continuous function’ is pivotal in understanding mathematical analysis and its applications. Imagine you’re drawing a line without lifting your pencil from the paper – intuitively, this is similar to a function being continuous. In the context of metric spaces, continuity has a precise definition that revolves around the idea that small changes in the input of the function result in small changes in the output.

A function \(f : (A, \rho) \rightarrow (B, \sigma)\) is said to be continuous at a point \(u \in A\) if for every \(\epsilon > 0\) there exists a \(\delta > 0\) such that, for all points \(v\) in \(A\), \(\rho(u, v) < \delta\) implies \(\sigma(f(u), f(v)) < \epsilon\). This tells us that as \(v\) stays within a \(\delta\)-neighborhood around \(u\), the image under \(f\), \(f(v)\), remains within an \(\epsilon\)-neighborhood around \(f(u)\).

The exercise provided showcases that if the functions \(f\) and \(g\) are continuous, and we follow the step-by-step solution, it becomes clear how the continuity of the composite function \(h = g \circ f\) is established.

Composition of Functions

The composition of functions is a process that combines two functions into a single function. Think of it like a relay race where the output of one runner (function) is passed on to the next. In mathematical terms, if \(f : A \rightarrow B\) and \(g : B \rightarrow C\) are functions, then the composition \(g \circ f\) is a function defined from \(A\) to \(C\) where for every \(u \in A\), \(g \circ f(u) = g(f(u))\).

For the composition to make sense, the range of the first function \(f\) must be compatible with the domain of the second function \(g\).

Confirming Continuity in Composition

As demonstrated in the original exercise, continuity of the composite function \(h = g \circ f\) on metric space \(A\) can be established by verifying the continuity condition for every point in \(A\). That is, we check that for every \(\epsilon > 0\), there exists some \(\delta > 0\) so the distance in \(C\) between \(h(u)\) and \(h(v)\) remains less than \(\epsilon\) whenever the distance in \(A\) between \(u\) and \(v\) is less than \(\delta\). The step-by-step solution to the exercise concluded that this condition holds, and thus, \(h\) maintains the chain of continuity from \(f\) to \(g\).