Consider the five topologies on \(\mathbb{R}\) given in Exercise 7 of §13.
Determine the closure of the set \(K = \ts{1/n \mid n \in \mathbb{Z}_+}\) under each of these topologies.
Which of these topologies satisfy the Hausdorff axiom? the \(T_1\) axiom?
Solution
The five topologies of E-ALTNF are: \(\mathcal T_1\) the standard topology; \(\mathcal T_2\) the topology of \(\mathbb R_K\), with basis the intervals \((a,b)\) and the sets \((a,b)-K\); \(\mathcal T_3\) the finite complement topology; \(\mathcal T_4\) the upper limit topology, with basis the intervals \((a,b]\); and \(\mathcal T_5\) the topology with basis the rays \((-\infty,a)\).
1. In \(\mathcal T_1\), \(\overline K=\boxed{K\cup\{0\}}\).
Every interval about \(0\) contains \(1/n\) for large \(n\), so \(0\in\overline K\). A point \(x\notin K\cup\{0\}\) has an interval about it missing \(K\): take \((-\infty,0)\) if \(x<0\), \((1,\infty)\) if \(x>1\), and \((\frac1{n+1},\frac1n)\) if \(\frac1{n+1}<x<\frac1n\).
2. In \(\mathcal T_2\), \(\overline K=\boxed{K}\).
1.So \(\mathbb R-K\) is open.
3. In \(\mathcal T_3\), \(\overline K=\boxed{\mathbb R}\).
The closed sets are \(\mathbb R\) and the finite sets, and \(K\) is infinite.
4. In \(\mathcal T_4\), \(\overline K=\boxed{K}\).
The basic set \((-1,0]\) contains \(0\) and misses \(K\). For \(x<0\), \(x>1\), and \(\frac1{n+1}<x<\frac1n\), the basic sets \((x-1,x]\), \((1,x]\), and \((\frac1{n+1},x]\) contain \(x\) and miss \(K\). So \(\mathbb R-K\) is open.
5. In \(\mathcal T_5\), \(\overline K=\boxed{[0,\infty)}\).
A union of rays \((-\infty,a)\) is \(\varnothing\), \(\mathbb R\), or a ray \((-\infty,c)\), so the closed sets are \(\varnothing\), \(\mathbb R\), and the rays \([c,\infty)\). A ray \([c,\infty)\) contains \(K\) if and only if \(c\le\frac1n\) for every \(n\), that is, \(c\le0\); the smallest is \([0,\infty)\).
6. \(\mathcal T_1\), \(\mathcal T_2\), and \(\mathcal T_4\) are Hausdorff, hence \(T_1\).
\(\mathcal T_1\) is the metric topology of \(\mathbb R\), and \(\mathcal T_2\) and \(\mathcal T_4\) are finer than \(\mathcal T_1\) by E-ALTNF; disjoint \(\mathcal T_1\)-open neighborhoods remain open in a finer topology.
7. \(\mathcal T_3\) is \(T_1\) but not Hausdorff.
Singletons are finite, hence closed. Two nonempty open sets have finite complements, so their intersection has finite complement in the infinite set \(\mathbb R\) and is nonempty.
8. \(\mathcal T_5\) is neither \(T_1\) nor Hausdorff.
By the description of closed sets in step 5, the closure of \(\{x\}\) is \([x,\infty)\ne\{x\}\). A Hausdorff space is \(T_1\).
9. Q.E.D.
Steps 1 through 5 answer (a), and steps 6 through 8 answer (b).