Question

Let \(0^{\prime}\) denote a point that is is not an element of \((-1,1)\), and let \(X=(-1,1) \cup\) \(\left\\{0^{\prime}\right\\}\). Let \(\mathcal{T}\) be the topology on \(X\) generated by the sets \((-1, a),(a, 1),[(-1, b) \backslash\) \(\\{0\\}] \cup\left\\{0^{\prime}\right\\}\), and \([(c, 1) \backslash\\{0\\}] \cup\left\\{0^{\prime}\right\\}\) where \(-1

Short answer

a. Both \(f\) and \(g\) are homeomorphisms onto their ranges. b. \(X\) is \(T_1\) but not Hausdorff. c. Given sets are compact but not closed, and their intersection isn't compact.

Step-by-step solution

Understand the Topology on X

The topology \( \mathcal{T} \) on \( X = (-1, 1) \cup \{0'\} \) includes open intervals and sets that exclude 0 but include \(0'\). It's important to see how \(0'\) is used to split \((-1,1)\) at 0, creating unique open sets around \(0'\).

Define Functions f and g

Define \( f : (-1,1) \rightarrow X \) by \( f(x) = x \) for all \( x \), making it the identity function on \((-1,1)\). Define \( g : (-1,1) \rightarrow X \) by \( g(x) = x \) for \( x eq 0 \) and \( g(0) = 0' \), mapping 0 to \(0'\). Both functions map points in \((-1,1)\) to their respective counterparts in \(X\).

Check Homeomorphisms for f and g

To show \( f \) is a homeomorphism, note it's bijective and continuous with the identity map's properties maintained in both \(X\) and the standard topology on \((-1,1)\). For \( g \), note that it switches 0 with \(0'\), and preserves continuity and bijection via the topology \(\mathcal{T}\) on \(X\), showing \( g \) is also a homeomorphism.

Analyze X's T1 but Non-Hausdorff Nature

In \(X\), any point except for 0 and 0' can be separated from others by open sets, showing it's \(T_1\). However, there are no disjoint open sets that can separate 0 and 0', which shows \(X\) is not Hausdorff.

Investigate Compactness and Closure of Sets

The set \([-\frac{1}{2}, \frac{1}{2}]\) is compact in the usual topology on \((-1,1)\). The set \(\left([-\frac{1}{2}, \frac{1}{2}] \backslash \{0\}\right) \cup \{0'\}\) is likewise compact under \(\mathcal{T}\) because \( \{0'\} \) behaves like \(0\) in terms of closure in compact sets. Both sets, however, are not closed in \(X\) under \(\mathcal{T}\) due to the neighborhood behavior of 0 and \(0'\). Their intersection lacks compactness, as the usual closure defining compactness isn't satisfied within \(X\).

Key concepts

Homeomorphism

In topology, a homeomorphism is a function that captures the concept of an "equivalent" shape or space. To be a homeomorphism, a function must be:

• Bijective: One-to-one and onto. Every element has a unique partner in the target set.
• Continuous: Small changes in input lead to small changes in output.
• Inverse Continuous: The reverse map is also continuous.

For the functions discussed in the exercise, like function \( f(x) = x \) for all \( x \) in \((-1, 1)\), it matches points directly between \((-1, 1)\) and set \( X \). This makes it a clear homeomorphism because it meets all three criteria above. Additionally, \( g(x) \) differs by mapping 0 to \( 0' \), but still maintains these conditions under the topology \( \mathcal{T} \). This means both \( f \) and \( g \) effectively preserve the topological structure of their domains, verifying their homeomorphic nature.

Hausdorff space

A Hausdorff space is a type of topological space where any two points can be separated by neighborhoods. This separation means there exist disjoint open sets around each point.In the presented problem, the space \( X \) is described as being \( T_1 \), meaning each point is closed by itself. Despite this, \( X \) is not a Hausdorff space because there are no disjoint open sets that separate the point at 0 from the point \( 0' \). This means even though they behave separately, they can't be distinctly "closed off" from each other in the topology \( \mathcal{T} \) on \( X \). Thus, while each point of \( X \) may have homeomorphic neighborhoods that pretend to act like a standard Hausdorff space, the whole space \( X \) does not satisfy the full Hausdorff requirement.

Compactness

Compactness can be thought of as a generalization of finiteness in a topological space. A set is compact if every open cover has a finite subcover. This means that we can cover the entire space using a few of these open sets.In the problem, we explored compactness within the sets \([-\frac{1}{2}, \frac{1}{2}]\) and \(([-\frac{1}{2}, \frac{1}{2}] \setminus \{0\}) \cup \{0'\}\). Both of these are compact within their respective topologies, however, they are not closed in the topology \( \mathcal{T} \) on \( X \). Interestingly, when these compact sets intersect, they lose compactness because the intersection does not behave like a standard compact set in \( X \). This shows the delicate balance required to maintain compactness and closure in different topological structures.

Topology on sets

The topology \( \mathcal{T} \) on a set determines the types of open sets that can exist. These open sets define how we understand concepts like continuity, limits, and compactness in a space.In the exercise, the topology \( \mathcal{T} \) on set \( X = (-1, 1) \cup \{0'\} \) is crafted using intervals that strategically exclude or include the point \( 0' \). This particular topology breaks typical intuitions by allowing \( 0' \) to create a unique structure around what is normally an interval \((-1, 1)\). By interpreting the topology \( \mathcal{T} \), one can see how classic properties like open sets and continuity adapt in this context. The topology \( \mathcal{T} \) helps to illustrate how small changes in set definitions can significantly impact the properties of those sets within a topological framework.