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Chapter 8: Problem 2

A metric space \((A, \rho)\) is connected if \(A\) cannot be written as \(A=A_{1} \cup A_{2},\) where \(A_{1}\) and \(A_{2}\) are nonempty disjoint open sets. Suppose that \((A, \rho)\) is connected and \(f:(A, \rho) \rightarrow(B, \sigma),\) where \(D_{f}=A, R_{f}=B,\) and \(f\) is continuous on \(A .\) Show that \((B, \sigma)\) is connected.

Short Answer

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Using the step by step solution above, answer the following question: **Question**: If \(f:(A, \rho) \rightarrow (B, \sigma)\) is a continuous function, and \((A, \rho)\) is a connected metric space, prove that \((B, \sigma)\) is connected. **Answer**: To prove that \((B, \sigma)\) is connected, suppose, for the sake of contradiction, that it is not connected. This means that there exist two nonempty disjoint open sets \(B_1\) and \(B_2\) in \((B, \sigma)\) such that \(B_1 \cup B_2 = B\) and \(B_1 \cap B_2 = \emptyset\). Since \(f\) is continuous, the inverse image of an open set in \((B, \sigma)\) is an open set in \((A, \rho)\). Let \(A_1 = f^{-1}(B_1)\) and \(A_2 = f^{-1}(B_2)\). Observe that \(A_1\) and \(A_2\) are open in \((A, \rho)\), nonempty, and disjoint. Moreover, \(A_1 \cup A_2 = A\) and \(A_1 \cap A_2 = \emptyset\). This contradicts the connectedness of \((A, \rho)\). Hence, assuming \((B, \sigma)\) is not connected leads to a contradiction, which means that \((B, \sigma)\) must be connected.

Step by step solution

01

Define connectedness in metric spaces

A metric space \((A, \rho)\) is connected if it cannot be written as a union of two nonempty disjoint open sets, i.e., there do not exist \(A_1, A_2 \subseteq A\) such that \(A_1 \cap A_2 = \emptyset\), \(A_1 \cup A_2 = A\), and both \(A_1\) and \(A_2\) are open in \((A, \rho)\).
02

Assume \((B, \sigma)\) is not connected

Let's suppose, for the sake of contradiction, that \((B, \sigma)\) is not connected. That means there exist two nonempty disjoint open sets \(B_1\) and \(B_2\) in \((B, \sigma)\) such that \(B_1 \cup B_2 = B\) and \(B_1 \cap B_2 = \emptyset\).
03

Use the properties of continuous functions to derive a contradiction

Since \(f: (A, \rho) \rightarrow (B, \sigma)\) is continuous, the inverse image of an open set in \((B, \sigma)\) is an open set in \((A, \rho)\). Let \(A_1 = f^{-1}(B_1)\) and \(A_2 = f^{-1}(B_2)\). Observe that \(A_1\) and \(A_2\) are open in \((A, \rho)\) because \(B_1\) and \(B_2\) are open in \((B, \sigma)\) and \(f\) is continuous. Moreover, \(A_1 \cup A_2 = f^{-1}(B_1 \cup B_2) = f^{-1}(B) = A\) and \(A_1 \cap A_2 = f^{-1}(B_1 \cap B_2) = f^{-1}(\emptyset) = \emptyset\). Note that \(A_1\) and \(A_2\) are nonempty, since \(B_1\) and \(B_2\) are nonempty and \(f\) is a function from \(A\) to \(B\). Thus, we have written \(A\) as a union of two nonempty disjoint open sets \(A_1\) and \(A_2\), contradicting the connectedness of \((A, \rho)\).
04

Conclusion

Since assuming \((B, \sigma)\) is not connected leads to a contradiction, we conclude that \((B, \sigma)\) must be connected.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuous Functions
A continuous function is one of the foundational concepts in mathematics. In simple terms, a function \( f: A \rightarrow B \) is considered continuous if small changes in the input (from space \(A\)) result in small changes in the output (to space \(B\)).
This can be visualized as a curve without any breaks, jumps, or holes.
This intuitive idea can be rigorously defined in terms of limits, but within metric spaces, continuity has a very useful property concerning open sets.

In the context of metric spaces, a function \( f: (A, \rho) \rightarrow (B, \sigma) \) is continuous if for every open set \( V \) in \( B \), the preimage \( f^{-1}(V) \) is an open set in \( A \). This means that the structure of openness is "preserved" in a way when mapped backward through the function. This property is pivotal when discussing connected spaces and stems from how continuity ensures no gaps or abrupt changes during the mapping.
Understanding continuity in terms of open sets is crucial, as it directly relates to how the spaces are interconnected or separated. This concept becomes particularly significant when discussing the connectedness of metric spaces through continuous maps.
Open Sets
Open sets are an essential component of topology, which is a branch of mathematics concerned with the properties of space that are preserved under continuous transformations. To intuitively understand open sets, think about them as spaces where, for every point within the set, you can move a little in any direction and still remain within the set.

In metric spaces, a set \( U \) is open if, for every point \( x \) in \( U \), there exists some radius \( \epsilon > 0 \) such that all points within the distance \( \epsilon \) from \( x \) are also in \( U \).
This implies that open sets do not include their boundary elements, providing a "cushioned" space without sharp edges. Open sets help define topological concepts like connectedness.

In connection with the problem of connected metric spaces, open sets are crucial because a disconnected space can be expressed as a union of two nonempty disjoint open sets. The definition hinges on the idea that no such separation into purely open disjoint sets is possible in a connected space.
Inverse Image
The inverse image (or preimage) of a set under a function is a concept that helps us track the points in the domain that get mapped to a particular subset of the codomain. For a function \( f: A \rightarrow B \) and a subset \( V \subseteq B \), the inverse image \( f^{-1}(V) \) is the set of all points in \( A \) that \( f \) maps into \( V \).

Mathematically, if \( f(x) \in V \), then \( x \in f^{-1}(V) \). This becomes very important when discussing continuous functions because, by definition, if \( V \) is an open set in \( B \), then \( f^{-1}(V) \) is an open set in \( A \) if \( f \) is continuous.
  • The inverse image allows us to study properties of \( B \) by looking at \( A \).
  • It helps relate the topology (open sets) of one space to another via a function.
  • In problems of connectedness, as explored in the exercise, the property of inverse image helps in building a contradiction when assuming non-connectedness.
This demonstrates the vital role inverse images play in preserving the structure of topological spaces when using continuous mappings.

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Most popular questions from this chapter

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 \(A\).

Show that if \(A\) is an arbitrary nonempty set, then $$ \rho(u, v)=\left\\{\begin{array}{ll} 0 & \text { if } v=u \\ 1 & \text { if } v \neq u \end{array}\right. $$ is a metric on \(A\).

Suppose that \(f:(A, \rho) \rightarrow(B, \sigma)\) and \(D_{f}=A\). Show that the following statements are equivalent. (a) \(f\) is continuous on \(A\). (b) If \(V\) is any open set in \((B, \sigma)\), then \(f^{-1}(V)\) is open in \((A, \rho)\). (c) If \(V\) is any closed set in \((B, \sigma)\), then \(f^{-1}(V)\) is closed in \((A, \rho)\).

Let \(C[0, \infty)\) be the set of all real-valued functions continuous on \([0, \infty)\). For each nonnegative integer \(n,\) let $$ \|f\|_{n}=\max \\{|f(x)| \mid 0 \leq x \leq n\\} $$ and $$ \rho_{n}(f, g)=\frac{\|f-g\|_{n}}{1+\|f-g\|_{n}} $$ Define $$ \rho(f, g)=\sum_{n=1}^{\infty} \frac{1}{2^{n-1}} \rho_{n}(f, g) $$ (a) Show that \(\rho\) is a metric on \(C[0, \infty)\). (b) Let \(\left\\{f_{k}\right\\}_{k=1}^{\infty}\) be a sequence of functions in \(C[0, \infty)\). Show that $$ \lim _{k \rightarrow \infty} f_{k}=f $$ in the sense of Definition 8.1 .14 if and only if $$ \lim _{k \rightarrow \infty} f_{k}(x)=f(x) $$ uniformly on every finite subinterval of \([0, \infty)\). (c) Show that \((C[0, \infty), \rho)\) is complete.

Suppose that \(\mathbf{X}=\left\\{x_{i}\right\\}_{i=1}^{\infty}\) is in \(\ell_{p},\) where \(p>1\). Show that (a) \(\mathbf{X} \in \ell_{r}\) for all \(r>p\) (b) If \(r>p,\) then \(\|\mathbf{X}\|_{r} \leq\|\mathbf{X}\|_{p}\) (c) \(\lim _{r \rightarrow \infty}\|\mathbf{X}\|_{r}=\|\mathbf{X}\|_{\infty}\).
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