Section 13: Problem 7 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Consider the following topologies on $\mathbb{R}$ : $\begin{eqnarray*} \mathcal{T}_{1} & = & \mbox{the standard topology},\\ \mathcal{T}_{2} & = & \mbox{the topology of }\mathbb{R}_{K},\\ \mathcal{T}_{3} & = & \mbox{the finite complement topology},\\ \mathcal{T}_{4} & = & \mbox{the upper limit topology, having all sets }(a,b]\mbox{ as basis},\\ \mathcal{T}_{5} & = & \mbox{the topology having all sets }(-\infty,a)=\{x|x<a\}\mbox{ as basis}. \end{eqnarray*}$ Determine, for each of these topologies, which of the others it contains.

Let " $\prec$ " be the "being smaller" relation on the set of topologies (as in Exercise 2). Then $3\prec1\prec2\prec4$ and $5\prec1\prec2\prec4$ , but $3$ and $5$ are not comparable. The relations are easy to see using Lemma 13.3. To see that $3$ and $5$ are not comparable consider point $1$ and two open sets $\mathbb{R}-\{0\}\in\mathcal{T}_{3}$ and $(-\infty,2)\in\mathcal{T}_{5}$ containing it. Neither of these open sets contains an open set from the other topology that contains $1$ .

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Section 13: Problem 7 Solution

Section 13: Problem 7 Solution