This is question 6 on page 83 in Chapter 13 of James R. Munkers’ Topology 2nd Edition textbook.

Proof

Setup

Given the topologies generated by \(\mathbb{R}_L\) and \(\mathbb{R}_K\), \(\mathcal{T}_L\) and \(\mathcal{T}_R\) respectively, show that \(\mathcal{T}_L\) and \(\mathcal{T}_R\) are not comparable.

This means showing \(\mathcal{T}_L \not \subset \mathcal{T}_K\) and \(\mathcal{T}_R \not \subset \mathcal{T}_K\).

Let $\mathcal{L}$ be basis for \(\mathbb{R}_L\) defined by \(\{x: a \le x \le b\) for all \(a, b\) in \(\mathbb{R}\}\).
Let $\mathcal{K}$ be basis for \(\mathbb{R}_K\) defined by \(\{x: x \in (a, b)\) or \(x \in (a, b) - K, K = \{\frac{1}{n}: n \in \mathbb{Z}_+\},\) for all \(a, b\) in \(\mathbb{R}\}\).

Show \(\mathcal{L} \not \subset \mathcal{K}\)

Need to show:

For some \(\mathcal{l} \in \mathcal{L}\), for some \(x \in \mathcal{l}\), there does not exist \(\mathcal{k} \in \mathcal{K}\) such that \(x \in \mathcal{k} \subset \mathcal{l}\).

For instance, suppose basis element \([x, d) \in \mathcal{L}\). There’s no open interval \((a, b) \in \mathcal{K}\) that contains \(x\) and lies in \([x, d)\). Therefore, \(\mathcal{L} \not \subset \mathcal{K}\).

Show \(\mathcal{K} \not \subset \mathcal{L}\)

Let \(B = (-1, 1) - K \in \mathcal{K}\). Choosing an arbitrary half-open interval in \(\mathcal{L}\), \(y = [-0.5, 0.5]\), \(B\) does not contain this interval as, in particular but could be applied to more than just this point, \(0.25\) is not in \(y\).

Thus, \(\mathcal{K} \not \subset \mathcal{L}\).

End of Proof

Because these two topologies are not subsets of each other, they are not comparable.

\[\tag*{$\blacksquare$}\]

Extras

link to the mathjax LaTeX specification: https://treeofmath.github.io/tex-commands-in-mathjax/TeXSyntax.htm