8.2 Examples[2BD]

E250

[0HM] Let’s consider on \(ℝ\) the family \(τ_+ =\{ (a, +∞) : a ∈ ℝ\} ∪ \{ ∅, ℝ\} \). Show that it is a topology. Is it Hausdorff? Calculate closure, interior, boundary and derivative of these sets:

\begin{align*} \{ 0\} \quad ,\quad \{ 0,1\} \quad ,\quad [0,1] \quad ,\quad (0,1) \quad ,\\[0,∞) \quad ,\quad (-∞,0] \quad ,\quad (0,∞) \quad ,\quad (-∞,0) \quad . \end{align*}

Hidden solution: [UNACCESSIBLE UUID ’0HN’]

E250

[0HP] Prerequisites:2,2.Let \(X=ℝ∪\{ +∞,-∞\} \), consider the family \(\mathcal B\) of parts of \(X\) that contains

  • open intervals \((a,b)\) with \(a,b∈ℝ\) and \(a{\lt}b\),

  • half-lines \((a,+∞]=(a,+∞)∪\{ +∞\} \) with \(a∈ℝ\),

  • the half-lines \([-∞,b)=(-∞,b)∪\{ -∞\} \) with \(b∈ℝ\).

(Note the similarity of sets in the second and third points with the ”neighbourhoods of infinity” seen in Sec. 6.1).

Show that \(\mathcal B\) satisfies the properties (a),(b) seen in 2. Let \(𝜏\) therefore be the topology generated from this base. The topological space \((X,𝜏)\) is called extended line, often denoted \(\overlineℝ\).

This topological space is \(T_ 2\), it is compact (Exercise 5), and is homoemorphic to the interval \([0,1]\). It can be equipped with a distance that generates the topology described above.

Hidden solution: [UNACCESSIBLE UUID ’0HQ’]

E250

[0HR] Prerequisites:2,2.Let \(X=ℝ∪\{ ∞\} \), let’s consider the family \(\mathcal B\) of parts of \(X\) comprised of

  • the open intervals \((a,b)\) with \(a,b∈ℝ\) and \(a{\lt}b\),

  • the sets \((a,+∞)∪(-∞,b)∪\{ ∞\} \) with \(a,b∈ℝ\) and \(a{\lt}b\).

Show that \(\mathcal B\) satisfies the properties (a),(b) seen in 2. Let \(𝜏\) therefore be the topology generated by this base. The topological space \((X,𝜏)\) is called one-point compactified line. This topological space is \(T_ 2\) and it is compact (Exer. 6); it is homeomorphic to the circle (Exer. 6); therefore it can be equipped with a distance that generates the topology described above.

E250

[0HS] Topics:directed ordering.Prerequisites:55.

Let \((J,≤)\) be a set with direct ordering. We decide that an ”open set” in \(J\) is a set \(A\) that contains a ”half-line” of the form \(\{ k∈ J : k≥ j\} \) (for a \(j∈ J\))  1 . Let therefore \(𝜏\) be the family of all such open sets, to which we add \(∅,J\). Show that \(𝜏\) is a topology. Is this topology Hausdorff? What are the accumulation points?

E250

[0HT] Topics:accumulation point, maximum, direct ordering.Prerequisites:55, 4.

Find a simple example of a set \((J,≤)\) with direct ordering that has maximum but, when we associate to \(J\) the topology \(𝜏_ J\) of the previous example, \((J,𝜏_ J)\) has no accumulation points.

Hidden solution: [UNACCESSIBLE UUID ’0HV’]

E250

[0HW] Topics:direct ordering. Prerequisites:53, 55, 3.

Let \((I,≤)\) be a set with direct ordering and with a maximum that we call \(∞\). We call \(J=I⧵\{ ∞\} \) and assume that \(J\) is filtering (with induced sorting) and non-empty. In this case we propose a finer topology. The topology \(𝜏\) for \(I\) contains:

  • \(∅,I\);

  • sets \(A\) that contain a ”half-line” \(\{ k∈ I : k≥ j\} \), for a \(j{\lt}∞\), (these are called “neighborhoods of \(∞\)”);

  • subsets of \(I\) that do not contain \(∞\).

Show that \(𝜏\) is a topology. Is this topology Hausdorff? Show that \(∞\) is the only accumulation point.

Hidden solution: [UNACCESSIBLE UUID ’0HX’]

The previous construction can be used in this way.

Remark 251

[0HY] Let \((J,≤)\) be a non-empty set with filtering order. We know from 3 that \(J\) has no maximum. We extend \((J,≤)\) by adding a point ”\(∞\)”: Let’s set \(I=J∪ \{ ∞\} \) and decide that \(x≤ ∞\) for every \(x∈ J\). It is easy to verify that \((I,≤)\) is a direct order, and obviously \(∞\) is the maximum \(I\).  2 Let \(𝜏\) be the topology defined in 6. We know that \(∞\) is an accumulation point. This topology can explain, in a topological sense, the limit already defined in 237, and other examples that we will see in Sec. 8.7.

[UNACCESSIBLE UUID ’0HZ’]

[UNACCESSIBLE UUID ’0J0’]

  1. We could call such a \(A\) a neighborhood of infinity, as was already done in Sec. 6.1.
  2. So \((I,≤)\) is not a filtering order.