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If \(X\) is an infinite set with the cofinite topology and \(\left\\{x_{j}\right\\}\) is a sequence of distinct points in \(X\), then \(x_{j} \rightarrow x\) for every \(x \in X\).

Short Answer

Expert verified
In cofinite topology, \(x_j \rightarrow x\) for every point \(x\) in \(X\) because the sequence eventually stays in every open set containing \(x\).

Step by step solution

01

Understand Cofinite Topology

In the cofinite topology on a set \(X\), the open sets are those whose complements in \(X\) are finite, along with the empty set. This means that any open set \(U\) contains all but finitely many elements of \(X\) (or it's the entire set \(X\)).
02

Define Sequence Convergence in Topological Terms

A sequence \(\{x_j\}\) converges to a point \(x\) in a topological space if, for every open set \(U\) containing \(x\), there exists a natural number \(N\) such that for all \(j \geq N\), \(x_j \in U\).
03

Apply Cofinite Topology to Sequence Convergence

Given the sequence \(\{x_j\}\) and any point \(x\) in \(X\), we need to examine the open sets that contain \(x\). In the cofinite topology, an open set \(U\) containing \(x\) must be such that its complement \(X \setminus U\) is finite. Therefore, \(\{x_j\}\) will eventually be inside \(U\) because there are only finitely many elements in \(X \setminus U\), and our sequence has distinct elements.
04

Conclude Convergence for Every Point \(x \in X\)

For any chosen point \(x \in X\), since the sequence eventually lies in every cofinite open set containing \(x\), the sequence \(\{x_j\}\) indeed converges to \(x\). Thus, \(x_j \rightarrow x\) for every \(x \in X\) under the cofinite topology.

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Key Concepts

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

Topological Space
A topological space is a fundamental concept in topology, a branch of mathematics that explores the properties of space that are preserved under continuous transformations. To create a topological space \(X\), you start with a set and a collection of subsets, known as open sets. This collection must satisfy three specific conditions:
  • The empty set and the entire set \(X\) must be open sets.
  • The union of any collection of open sets is also an open set.
  • The intersection of any finite number of open sets is an open set.
These rules ensure a structure where concepts like continuity, convergence, and neighborhood can be defined. In the exercise, \(X\) is an infinite set endowed with the cofinite topology. This particular topology has its defined set of open sets, making the exercise a great example of applying these theoretical conditions in a specific context.
Sequence Convergence
In the context of topology, sequence convergence generalizes the idea from calculus, but it is not always about approaching a limit number-wise. In a topological space, a sequence \(\{x_j\}\) converges to a point \(x\) if, for every open set \(U\) containing \(x\), there's a natural number \(N\) such that for all terms \(x_j\) of the sequence with indices \(j\geq N\), the terms are inside \(U\).

This means that farther along the sequence, the points \(x_j\) get arbitrarily close to the point \(x\). In our exercise, this nuance of convergence assures that given any open set (with specific properties due to the cofinite topology), past a certain point in the sequence, all terms are contained within this open set.

Convergence, in this sense, becomes a tool to discuss the layout and behavior of sequences in a broader sense, beyond mere proximity in a typical numerical sense.
Open Sets
Open sets are central to understanding topologies, acting like loose 'containers' in a space. They help define various topological properties and are essential to many foundational concepts.

In the cofinite topology, a distinctive type of topology used in the problem, open sets are defined as those whose complements in the set are finite. Beyond the empty set, any open set \(U\) in this topology includes all but finitely many elements of the entire set. This structure makes it particularly suitable when dealing with infinite sets like \(X\).
  • In essence, except for a limited number of exceptions (which are finite), every point in \(X\) belongs to any open set containing \(x\).
  • This contributes to discussions on sequence convergence, as it guarantees that sequences with sufficiently high index values will belong to any open set containing the limit point, thus qualifying as convergent in this topological sense.
Given this backdrop, cofinite open sets allow for elegant solutions and the examination of sequence behavior in infinite spaces.

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Most popular questions from this chapter

An open cover \(\mathcal{U}\) of a topological space \(X\) is called locally finite if each \(x \in X\) has a neighborhood that intersects only finitely many members of \(U\). If \(U, \mathcal{V}\) are open covers of \(X, V\) is a refinement of \(U\) if for each \(V \in V\) there exists \(U \in \mathcal{U}\) with \(V \subset U . X\) is called paracompact if every open cover of \(X\) has a locally finite refinement. a. If \(X\) is a \(\sigma\)-compact LCH space, then \(X\) is paracompact. In fact, every open cover \(U\) has locally finite refinements \(\left\\{V_{\alpha}\right\\},\left\\{W_{\alpha}\right\\}\) such that \(\bar{V}_{\alpha}\) is compact and \(\bar{W}_{\alpha} \subset V_{\alpha}\) for all \(\alpha\). (Let \(\left\\{U_{n}\right\\}_{1}^{\infty}\) be as in Proposition 4.39. For each \(n\), \(\left\\{E \cap\left(U_{n+2} \backslash \bar{U}_{n-1}\right): E \subset U\right\\}\) is an open cover of \(\bar{U}_{n+1} \backslash U_{n}\). Choose a finite

Let \(X\) be a topological space, \(Y\) a Hausdorff space, and \(f, g\) continuous maps from \(X\) to \(Y\). a. \(\\{x: f(x)=g(x)\\}\) is closed. b. If \(f=g\) on a dense subset of \(X\), then \(f=g\) on all of \(X\).

A topological space \(X\) is called disconnected if there exist nonempty open sets \(U, V\) such that \(U \cap V=\varnothing\) and \(U \cup V=X\); otherwise \(X\) is connected. When we speak of connected or disconnected subsets of \(X\), we refer to the relative topology on them. a. \(X\) is connected iff \(\varnothing\) and \(X\) are the only subsets of \(X\) that are both open and closed. b. If \(\left\\{E_{\alpha}\right\\}_{\alpha \in A}\) is a collection of connected subsets of \(X\) such that \(\bigcap_{\alpha \in A} E_{\alpha} \neq\) \(\varnothing\), then \(\bigcup_{\alpha \in A} E_{\alpha}\) is connected. c. If \(A \subset X\) is connected, then \(\bar{A}\) is connected. d. Every point \(x \in X\) is contained in a unique maximal connected subset of \(X\), and this subset is closed. (It is called the connected component of \(x\).)

If \(\left\\{X_{\alpha}\right\\}_{\alpha} \in A\) is a family of topological spaces of which infinitely many are noncompact, then every closed compact subset of \(\prod_{\alpha \in A} X_{\alpha}\) is nowhere dense.

Consider N (with the discrete topology) as a subset of its Stone-Cech compactification \(\beta N\). a. If \(A\) and \(B\) are disjoint subsets of N, their closures in \(\beta N\) are disjoint. (Hint: \(\left.X_{A} \in C(\mathrm{~N}, I) .\right)\) b. No sequence in N converges in \(\beta N\) unless it is eventually constant (so \(\beta N\) is emphatically not sequentially compact).

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