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Prove that if \(X\) and \(Y\) are connected, so is \(X \times Y\) under the product metric.

Short Answer

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The product \(X \times Y\) is connected if both \(X\) and \(Y\) are connected.

Step by step solution

01

Understanding Concepts

To prove that the product of two connected spaces, \(X\) and \(Y\), is also connected, it helps to first understand the definitions involved. A topological space is connected if it cannot be partitioned into two nonempty open sets that are disjoint from each other. The product of two spaces \(X \times Y\) has the product topology, where open sets are made from the basis consisting of products of open sets in \(X\) and \(Y\).
02

Assume the Contrary

Assume, for contradiction, that \(X \times Y\) is not connected. This means there exist nonempty open sets \(U\) and \(V\) in \(X \times Y\) such that \((X \times Y) = U \cup V\) and \(U \cap V = \emptyset\).
03

Expression of Open Sets in Product Topology

In the product topology, open sets \(U\) and \(V\) are unions of sets of the form \((U_1 \times V_1)\), where \(U_1\) is open in \(X\) and \(V_1\) is open in \(Y\). Therefore, if \(U\) and \(V\) partition \(X \times Y\), it imposes a partition on \(X\) or \(Y\).
04

Projection and Induced Partitions

Consider the projection functions \(p_X: X\times Y \rightarrow X\) and \(p_Y: X \times Y \rightarrow Y\). Since \(p_X(U) \cup p_X(V) = X\) and their intersection is empty, this implies \(X\) can be similarly parted. The same argument applies to \(Y\).
05

Contradiction and Conclusion

The projections \(p_X(U)\) and \(p_X(V)\) are open and disjoint sets that cover \(X\). However, this contradicts \(X\) being connected as \(X\) cannot be divided into two non-empty disjoint open sets. The same holds for \(Y\). Thus, our assumption is false, and \(X \times Y\) must be connected.

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Key Concepts

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

Topology
Topology is an area of mathematics that looks at the properties of space that are preserved under continuous transformations. This exciting field dives into concepts like convergence, compactness, and connectedness, and helps to formalize the ideas of shapes and spaces.

A topological space is essentially a set equipped with a topology, which is a collection of open sets. These open sets satisfy specific properties:
  • The empty set and the entire space must be included.
  • The intersection of a finite number of open sets is also an open set.
  • The union of any collection of open sets is an open set.
You can think of topological spaces as generalizations of geometric concepts. This allows mathematicians to talk about spaces without requiring rigid structure, like that found in metric spaces. Thus, topology embraces a more free-form view of why certain properties hold true across different types of spaces.
Product Topology
The product topology involves creating a new topological space from two other spaces. If you have spaces, say \(X\) and \(Y\), their product topology results in \(X \times Y\). This space contains pairs of elements, where the first is a member of \(X\) and the second belongs to \(Y\).

Open sets in the product topology are formed from the basis elements of the form \(U \times V\), where \(U\) is an open set in \(X\) and \(V\) is an open set in \(Y\). The collection of all such products forms the product topology's basis, bridging the structures of both spaces into one.

This concept is powerful because it allows the combination of diverse spaces into a singular framework. This also helps in analyzing important properties, like continuity and compactness, in a more complex structure consisting of simpler spaces.
Connectedness
Connectedness is a vital concept in topology that describes spaces that can't be split into two separate open sets. It defines what it means for a space to be 'in one piece.'

A space is connected if there's no way to divide it into two disjoint non-empty open sets. Anytime you assume such a division and it leads to contradictions—as seen in the exercise with spaces \(X\) and \(Y\) forming \(X \times Y\)—you confirm the space's connectedness.

To visualize, imagine a rubber band. If you can't sever it into two pieces without cutting, it remains as one connected object. This analogy extends to more intricate structures in topology, helping us explore properties maintaining continuity and cohesion.

In the context of product topology, understanding connectedness helps prove that combining two connected spaces \(X\) and \(Y\) means their product space \(X \times Y\) is also connected. This showcases how properties from individual spaces can influence the structure of a more complex, unified space.

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Most popular questions from this chapter

Let \(\bar{p}_{0}, \bar{p}_{1}, \ldots, \bar{p}_{m}\) be fixed points in \(E^{n}\left({ }^{*}\right.\) or in another normed space \()\). Let $$f(t)=\bar{p}_{k}+(t-k)\left(\bar{p}_{k+1}-\bar{p}_{k}\right)$$ whenever \(k \leq t \leq k+1, t \in E^{1}, k=0,1, \ldots, m-1\) Show that this defines a uniformly continuous mapping \(f\) of the interval \([0, m] \subseteq E^{1}\) onto the "polygon" $$\bigcup_{k=0}^{m-1} L\left[p_{k}, p_{k+1}\right]$$ In what case is \(f\) one to one? Is \(f^{-1}\) uniformly continuous on each \(L\left[p_{k}, p_{k+1}\right] ?\) On the entire polygon?

Prove that if the functions \(f_{n}\) and \(g_{n}\) are real or complex (or if the \(g_{n}\) are vector valued and the \(f_{n}\) are scalar valued), and if $$ f_{n} \rightarrow f \text { and } g_{n} \rightarrow g \text { (uniformly) on } B \text { , } $$ then $$ f_{n} g_{n} \rightarrow f g \text { (uniformly) on } B $$ provided that either \(f\) and \(g\) or the \(f_{n}\) and \(g_{n}\) are bounded on \(B\) (at least from some \(n\) onward \() ;\) cf. Problem 11 . Disprove it for the case where only one of \(f\) and \(g\) is bounded.

Show that the arcsine function is uniformly continuous on [-1,1] .

Verify that any infinite set in a discrete space is closed and bounded but not compact. [Hint: In such a space no sequence of distinct terms clusters.]

A set is said to be totally disconnected iff its only connected subsets are one-point sets and \(\emptyset\). Show that \(R\) (the rationals) has this property in \(E^{1}\).

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