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Suppose that \((X, \mathcal{J})\) is a compact Hausdorff space and \(\mathcal{T}^{\prime}\) is another topology on \(X .\) If \({ }^{\prime}{ }^{\prime}\) is strictly stronger than \({\mathcal{J}}\), then \(\left(X, \mathcal{J}^{\prime}\right)\) is Hausdorff but not compact. If \(\mathcal{J}^{\prime}\) is strictly weaker than \(\mathcal{J}\), then \(\left(X, \mathcal{T}^{\prime}\right)\) is compact but not Hausdorff.

Short Answer

Expert verified
A stronger topology keeps Hausdorff but may not be compact; a weaker topology keeps compact but may not be Hausdorff.

Step by step solution

01

Understanding the Problem

We have a compact Hausdorff space \((X, \mathcal{J})\). We need to analyze the implications when another topology \(\mathcal{T}^{\prime}\) on \(X\) is either strictly stronger or strictly weaker than \(\mathcal{J}\).
02

Stronger Topology and Compactness

If \(\mathcal{T}^{\prime}\) is strictly stronger than \(\mathcal{J}\), it means \(\mathcal{J}\) \(\subsetneq\) \(\mathcal{T}^{\prime}\). This can destroy compactness because compactness requires every open cover to have a finite subcover, and adding more open sets may mean more covers that do not have finite subcovers. Therefore, \((X, \mathcal{T}^{\prime})\) may not be compact.
03

Stronger Topology and Hausdorff Property

If \(\mathcal{T}^{\prime}\) is stronger, it maintains or enhances the Hausdorff property. In a stronger topology, any two distinct points can still be separated by disjoint open sets, possibly even more so than in \(\mathcal{J}\). Thus, \((X, \mathcal{T}^{\prime})\) remains Hausdorff.
04

Weaker Topology and Compactness

If \(\mathcal{T}^{\prime}\) is strictly weaker than \(\mathcal{J}\), it means \(\mathcal{T}^{\prime}\) \(\subsetneq\) \(\mathcal{J}\). This generally preserves compactness because fewer open sets mean fewer covers to consider, making it easier to have a finite subcover. Thus, \((X, \mathcal{T}^{\prime})\) is compact.
05

Weaker Topology and Hausdorff Property

When \(\mathcal{T}^{\prime}\) is weaker, it sometimes compromises the Hausdorff property because there may not be enough open sets to separate points that could previously be separated under \(\mathcal{J}\). Thus, \((X, \mathcal{T}^{\prime})\) may not be Hausdorff.

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Key Concepts

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

Compactness
Compactness is a fundamental concept in topology that deals with the structure of spaces. A topological space - is said to be compact if every open cover has a finite subcover. This means if you take a whole bunch of open sets that completely cover the space, you can still cover the space with just a finite number of these sets.

In simpler terms, compactness can be thought of like having a space that doesn't "stretch out" indefinitely – it can be covered by a limited number of small "blankets," no matter how many you originally had.
Understanding compactness is important because compact spaces behave nicely under continuous mappings, and they possess many useful properties. In the given exercise, it is noted that adding more open sets (making a stronger topology) might disrupt having finite subcovers since there could be more options to consider, thus destroying compactness.
Hausdorff Space
A Hausdorff space, also known as a "T2 space," is characterized by its distinct separation property:
- If two points are in a Hausdorff space, there exist open sets that separate them.

This means if you pick any two different points in the space, you can find two disjoint open sets that each contain one of the points and do not overlap.

It's like having enough flexibility to place individuals on different "floors" without them bumping into each other.
The exercise highlights that strengthening a topology (adding more open sets) tends to maintain or enhance the Hausdorff property because more distinct open sets might exist to effect the separation of points. However, weakening a topology might erode this capability, as there could be fewer open sets available for separation.
Topology Strength
Topology strength refers to how many open sets are part of a topology. - A stronger topology has more open sets compared to another topology on the same set. - Conversely, a weaker topology has fewer open sets. Understanding the strength of a topology helps in analyzing how properties such as compactness and the Hausdorff condition are affected.

In the exercise, when a topology is strictly stronger than another (more open sets), it might mess with the compactness since more open sets might create infinite cover challenges but maintain Hausdorffness.
On the flip side, a weaker topology (fewer open sets) generally keeps compactness intact because it simplifies finite covering but risks losing the ability to sufficiently separate points.

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Most popular questions from this chapter

Let \(\left\\{\left(X_{n}, \rho_{n}\right)\right\\}_{1}^{\infty}\) be a countable family of metric spaces whose metrics take values in \([0,1]\). (The latter restriction can always be satisfied; see Exercise \(56 \mathrm{~b}\).) Let \(X=\prod_{1}^{\infty} X_{n .}\) If \(x, y \in X\), say \(x=\left(x_{1}, x_{2}, \ldots\right)\) and \(y=\left(y_{1}, y_{2}, \ldots\right)\), define \(\rho(x, y)=\sum_{1}^{\infty} 2^{-n} \rho_{n}\left(x_{3}, y_{n}\right) .\) Then \(\rho\) is a metric that defines the product topology on \(X\).

Let \(X\) be a compact Hausdorff space. An ideal in \(C(X, \mathbb{R})\) is a subalgebra \(\mathcal{J}\) of \(C(X, \mathbb{R})\) such that if \(f \in J\) and \(g \in C(X, \mathbb{R})\) then \(f g \in J\). a. If \(J\) is an ideal in \(C(X, \mathbb{R})\), let \(h(\mathcal{J})=\\{x \in X: f(x)=0\) for all \(f \in \mathcal{J}\\} .\) Then \(h(\mathcal{J})\) is a closed subset of \(X\), called the hull of J. b. If \(E \subset X\), let \(k(E)=\\{f \in C(X, \mathbb{R}): f(x)=0\) for all \(x \in E\\}\). Then \(k(E)\) is a closed ideal in \(C(X, \mathbb{R})\), called the kernel of \(E\). c. If \(E \subset X\), then \(h(k(E))=\bar{E}\). d. If \(J\) is an ideal in \(C(X, R)\), then \(k(h(\mathcal{J}))=J\). ( Hint: \(k(h(J))\) may be identified with a subalgebra of \(C_{0}(U, \mathbb{R})\) where \(\left.U=X \backslash h(J) .\right)\) e. The closed subsets of \(X\) are in one-to-one correspondence with the closed ideals of \(C(X, \mathbb{R})\).

The one-point compactification of \(\mathbb{R}^{n}\) is homeomorphic to the \(n\)-sphere \(\\{x \in\) \(\left.\mathbb{R}^{n+1}:|x|=1\right\\}\).

Let \(1-\sum_{1}^{\infty} c_{n} t^{n}\) be the Maclaurin series for \((1-t)^{1 / 2}\). a. The series converges absolutely and uniformly on compact subsets of \((-1,1)\), as does the termwise differentiated series \(-\sum_{1}^{\infty} n c_{n} t^{n-1}\). Thus, if \(f(t)=\) \(1-\sum_{1}^{\infty} c_{n} t^{n}\), then \(f^{\prime}(t)=-\sum_{1}^{\infty} n c_{n} t^{n-1}\). b. By explicit calculation, \(f(t)=-2(1-t) f^{\prime}(t)\). from which it follows that \((1-t)^{-1 / 2} f(t)\) is constant. Since \(f(0)=1, f(t)=(1-t)^{1 / 2}\).

If \(A\) is a directed set, a subset \(B\) of \(A\) is called cofinal in \(A\) if for each \(\alpha \in A\) there exists \(\beta \in B\) such that \(\beta \gtrsim \alpha\). a. If \(B\) is cofinal in \(A\) and \(\left\langle x_{\alpha}\right\rangle_{\alpha \in A}\) is a net, the inclusion map \(B \rightarrow A\) makes \(\left\langle x_{\beta}\right\rangle_{\beta \in B}\) a subnet of \(\left\langle x_{\alpha}\right\rangle_{\alpha \in A}\). b. If \(\left\langle x_{\alpha}\right\rangle_{\alpha} \in A\) is a net in a topological space, then \(\left\langle x_{\alpha}\right\rangle\) converges to \(x\) iff for every cofinal \(B \subset A\) there is a cofinal \(C \subset B\) such that \(\left\langle x_{\gamma}\right\rangle_{\gamma \in C}\) converges to \(x\).

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