Consider the following topologies on \(\mathbb{R}\):
\(\mathcal{T}_1\) = the standard topology,
\(\mathcal{T}_2\) = the topology of \(\mathbb{R}_K\),
\(\mathcal{T}_3\) = the finite complement topology,
\(\mathcal{T}_4\) = the upper limit topology, having all sets \((a, b]\) as basis,
\(\mathcal{T}_5\) = the topology having all sets \((-\infty, a) = \ts{x \mid x < a}\) as basis.
Determine, for each of these topologies, which of the others it contains.
Solution
Write \(K=\{1/n : n\in\mathbb Z_+\}\), so that \(\mathcal T_2\) has basis the open intervals \((a,b)\) and the sets \((a,b)\setminus K\).
1. \(\mathcal T_3\subseteq\mathcal T_1\) and \(\mathcal T_5\subseteq\mathcal T_1\).
A nonempty \(\mathcal T_3\)-open set is \(\mathbb R\setminus\{x_1,\ldots,x_k\}=\bigcap_i(\mathbb R\setminus\{x_i\})\), a finite intersection of standard open sets. A basis element of \(\mathcal T_5\) is \((-\infty,a)=\bigcup_{n\ge1}(a-n,a)\).
2. \(\mathcal T_1\subseteq\mathcal T_2\subseteq\mathcal T_4\).
The basis of \(\mathcal T_2\) contains every open interval, so \(\mathcal T_1\subseteq\mathcal T_2\). Each \((a,b)=\bigcup_{n}(a,b-1/n]\) is \(\mathcal T_4\)-open. For \((a,b)\setminus K\), let \(x\) be a point of it. If \(x\le0\), then \((a,x]\subseteq(a,b)\setminus K\). If \(x>1\), then \((\max(a,1),x]\subseteq(a,b)\setminus K\). If \(0<x<1\), choose \(n\) with \(\frac1{n+1}<x<\frac1n\); then \((\max(a,\frac1{n+1}),x]\subseteq(a,b)\setminus K\). So \((a,b)\setminus K\) is \(\mathcal T_4\)-open, and \(\mathcal T_2\subseteq\mathcal T_4\).
3. \(\mathcal T_3\) and \(\mathcal T_5\) are incomparable.
The ray \((-\infty,0)\in\mathcal T_5\) has infinite complement, so it is not in \(\mathcal T_3\). \(\mathbb R\setminus\{0\}\in\mathcal T_3\), while every nonempty union of rays \((-\infty,a)\) is \(\mathbb R\) or a ray \((-\infty,c)\), so \(\mathbb R\setminus\{0\}\notin\mathcal T_5\).
4. \(\mathcal T_1\not\subseteq\mathcal T_3\), \(\mathcal T_1\not\subseteq\mathcal T_5\), \(\mathcal T_2\not\subseteq\mathcal T_1\), and \(\mathcal T_4\not\subseteq\mathcal T_2\).
\((0,1)\in\mathcal T_1\) has infinite complement and is not a union of rays. \((-1,1)\setminus K\in\mathcal T_2\) is not standard-open, since every interval around \(0\) meets \(K\). \((0,1]\in\mathcal T_4\) is not \(\mathcal T_2\)-open: a basis element of \(\mathcal T_2\) containing \(1\in K\) is an interval \((c,d)\) with \(d>1\), which is not contained in \((0,1]\).
5. Q.E.D.
By steps 1 and 2, \(\mathcal T_1\) contains \(\mathcal T_3\) and \(\mathcal T_5\); \(\mathcal T_2\) contains \(\mathcal T_1,\mathcal T_3,\mathcal T_5\); \(\mathcal T_4\) contains \(\mathcal T_1,\mathcal T_2,\mathcal T_3,\mathcal T_5\). By steps 3 and 4, \(\mathcal T_3\) and \(\mathcal T_5\) contain none of the others, and no further inclusion holds: \(\mathcal T_2,\mathcal T_4\supseteq\mathcal T_1\) are not contained in \(\mathcal T_3\) or \(\mathcal T_5\), \(\mathcal T_4\supseteq\mathcal T_2\) is not contained in \(\mathcal T_1\), and \(\mathcal T_4\not\subseteq\mathcal T_2\).