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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.

Srivatsan
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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
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    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
  • @Theo I thought the title was good. But feel free to make improvements! I just want to emphasize that my main trouble is in showing that this is a topology; I can show the remaining subparts. (May be we will soon have complete solutions for all exercises from that book in this site :-)) – Srivatsan Sep 18 '11 at 17:42
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    Oh, I was asking if the title of the book is accurate... – t.b. Sep 18 '11 at 17:43

1 Answers1

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$C = \bigcap\limits_{i\in I} (A_i \cup T_i)$, 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
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Brian M. Scott
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  • Thanks! Very nice approaches, both of them. (I'll upvote this tonight. :-/) – Srivatsan Sep 18 '11 at 19:38
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    @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 00:14
  • @BrianM.Scott Could you please give a bit more of a hint? I've been following the problem, but I don't think I understand what should be seen from $C\A$. In that collection of intersections, I think one of the terms will be exactly the term $A$ we're "differencing" out (I see this for |I|=2). I just don't understand what is left and how this implies the intersection is closed. Thanks :) – mathmath8128 Sep 19 '11 at 01:18
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    @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 01:49
  • @BrianM.Scott Thank you! I don't think I would've thought to consider the difference like that. I really appreciate your quick response. – mathmath8128 Sep 19 '11 at 01:54
  • @BrianM.Scott Could you give some hint for the "Hausdorff but not regular" part. I got stuck there. I tried the set like $\mathcal{S}\cup{0}$ or $\mathcal{S}\cup \left[0,\dfrac{1}{2}\right]$ but it seems not working. – Jebei Aug 09 '13 at 12:44
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    @frame99: You can't separate the point $0$ from the closed set $S$ with disjoint open sets. – Brian M. Scott Aug 09 '13 at 12:50
  • @brianM.Scott Does this problem similar to U[0-1/n,1]=(0,1]? – Jebei Aug 09 '13 at 13:17
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    @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 09 '13 at 13:29
  • @BrianM.Scott Hausdorff part is same as the usual topology on R？ – Jebei Aug 09 '13 at 13:33
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    @frame99: The topology is finer than the usual topology on $\Bbb R$, so it's automatically Hausdorff. – Brian M. Scott Aug 09 '13 at 13:38
  • @BrianM.Scott Thank you for your patient explaintion.I easily get confused with the definition for topology. – Jebei Aug 09 '13 at 14:12
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    @frame99: You're welcome. – Brian M. Scott Aug 09 '13 at 14:16
  • @BrianM.Scott sir i have some confusion in ur another approach answer .....Im thinking about ur answer like this $B(x, n) = { y \in \mathbb{R} : |y-x| < \frac{1}{2^n} } $ is open when $x \neq 0$,

    $B(x, n) = { y \in \mathbb{R} : |y-x| < \frac{1}{2^n} }- {S} $ is closed when $x = 0$

    Is its True/false , i mean is my thinking is true or not ??

    Another confusion about symbolics in $\mathscr{B}$ . what is the meaning of this symbol $\omega \land x ?$

     – jasmine Jul 18 '20 at 05:28
  • @jasmine: $B(x,n)={y\in\Bbb R:|y-x|<2^{-n}}$ is indeed open when $x\ne 0$, but $B(x,n)={y\in\Bbb R:|y-x|<2^{-n}}\setminus S$ is also open when $x=0$. That is, all of the sets $B(x,n)$ that I defined are open. The symbol $\land$ is the usual symbol for logical OR, so $n\in\omega\land x\in\Bbb R$ means the same as $n\in\omega\text{ and }x\in\Bbb R$. In other words, $n$ can be any natural number (including $0$), and $x$ can be any real number. – Brian M. Scott Jul 18 '20 at 05:44
  • okss that mean its about point $0$ , that is there exist no disjoint neighbourhood. thanks u sir @BrianM.Scott – jasmine Jul 18 '20 at 05:51