# Verifying that these sets form a topology

I am solving Exercise 4.1, Question 17(v) from Topology without Tears (link) by Sidney Morris. (This exercise is marked with a star.)

Let $S = \{ \frac{1}{n} \,:\, n \in \mathbb N \}$. Define a set $C \subseteq \mathbb R$ to be closed if $C = A \cup T$ where $A$ is closed in the euclidean topology on $\mathbb R$ and $T$ is any subset of $S$. [Show that] The complements of these closed sets form a topology $\mathcal T$ on $\mathbb R$ which is Hausdorff but not regular.

We need to show three things here: $\mathcal T\$ is a topology, it is Hausdorff, but it is not regular. I can prove the "Hausdorff but not regular" part, but I am confused about showing that this is actually a topology!

Here is my attempt. (I will put quotes around the words open and closed, whenever they are with respect to $\mathcal T\$, to remind myself that I haven't yet verified that $\mathcal T\$ is a topology.) Clearly I should just verify that $\mathcal T\$ satisfies the topology axioms.

1. I can see why both $\mathbb R$ and $\emptyset$ are "closed" sets in $\mathcal T\$.

2. I can also show that if $C_1 = A_1 \cup T_1$ and $C_2 = A_2 \cup T_2$ are two "closed" sets, then their union is also "closed". Hence, by induction, a finite union of "closed" sets is also "closed".

3. The part that I am stuck in is showing that an arbitrary intersection of "closed" sets is also closed. Let $I$ be an arbitrary index set. For $i \in I$, let $C_i = A_i \cup T_i$ be such that $A_i$ is closed in $\mathbb R$ and $T_i \subseteq S$. I want to write the intersection

$$C := \bigcap_{i \in I} (A_i \cup T_i)$$ in a form which makes it evident that $C$ is "closed". But naively distributing the $\bigcap$ over the $\cup$ does not seem to work. Please suggest some hints!

Though this is not really homework, I'll add the homework tag since I am not looking for complete solutions anyway.

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Try looking at the union of open sets instead of the intersection of closed ones. An open set in $\mathcal{T}$ is of the form $U \setminus T$, where $U$ is an open subset of $\mathbb R$ and $T$ is a subset of $S$. An arbitrary union of sets of the form $U_i \setminus T_i$ is the union of the $U_i$ (which is open because the $U_i$ are open) minus the intersection of the $T_i$ (which is a subset of S). – user15464 Sep 18 '11 at 17:11
@user15464 I did think about open sets as well and I got up to the criterion for open sets you mentioned. But whatever I could do with the definition of open sets, it seemed that I could also do with closed sets and vice versa. So let me try to translate your final statement in terms of closed sets: An arbitrary intersection of sets of the form $A_i \cup T_i$ is $\bigcap A_i$ (which is closed in $\mathbb R$) union $\bigcap T_i$ (which is a subset of $S$). I think this claim is clearly false. I am not sure about your claim; I'll think about it for some time and get back :-). – Srivatsan Sep 18 '11 at 17:20
@user15464 This is what I get: $$(V_1 \setminus T_1) \cup (V_2 \setminus T_2) = [(V_1 \cup V_2) \setminus (T_1 \cap T_2)] \cap (V_1 \cup T_2^c) \cap (V_2 \cup T_1^c).$$ (I changed the $U$'s to $V$s for ease of reading.) – Srivatsan Sep 18 '11 at 17:32
I seem to see an awful lot of questions from Topology without Tears here recently... Is the title really accurate? :) – t.b. Sep 18 '11 at 17:38
Oh, I was asking if the title of the book is accurate... – t.b. Sep 18 '11 at 17:43

$C = \bigcap\limits_{i\in I} (A_i \cup T_i)$, where each $A_i$ is Euclidean-closed, and each $T_i \subseteq S$. Let $A = \bigcap\limits_{i\in I} A_i$; certainly $A$ is Euclidean-closed. Where must any point of $C \setminus A$ be?
Another approach is via local bases. For $n \in \omega$ and $x \in \mathbb{R}$ let $$B(x,n) = \begin{cases} \{y \in \mathbb{R}:\vert y-x\vert < 2^{-n}\},&\text{if }x \ne 0\\ \{y \in \mathbb{R}:\vert y-x\vert < 2^{-n}\}\setminus S,&\text{if }x = 0. \end{cases}$$
Show that $\mathscr{B} = \{B(x,n):n \in \omega \land x \in \mathbb{R}\}$ is a base for a topology, and that the topology that it generates is $\mathcal{T}$. (Thus, $\mathcal{T}$ differs from the Euclidean topology only at $0$.)
@Srivatsan: You’re right about the typo: I was originally going to define $B(x,n)$ in-line, separately for the two cases, and forgot to change the lead-in when I shifted to the case display. – Brian M. Scott Sep 19 '11 at 0:14
@aengle: Suppose that $x\in C\setminus A$. Since $x\notin A$, there is at least one $i\in I$ s.t. $x\notin A_i$. But $x\in C$, so $x\in A_i\cup T_i$, and therefore $x\in T_i\subseteq S$. It follows that $C\setminus A\subseteq S$, and since $A$ is Euclidean-closed, $C=A\cup (C\setminus A)\in\mathcal{T}$. – Brian M. Scott Sep 19 '11 at 1:49
@frame99: You can't separate the point $0$ from the closed set $S$ with disjoint open sets. – Brian M. Scott Aug 9 '13 at 12:50
@frame99: Not really, no. It's just a matter of checking that if $U$ is an open nbhd of $0$, and $V$ is an open nbhd of $S$, then $U\cap V\ne\varnothing$. – Brian M. Scott Aug 9 '13 at 13:29
@frame99: The topology is finer than the usual topology on $\Bbb R$, so it's automatically Hausdorff. – Brian M. Scott Aug 9 '13 at 13:38