Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 4 down vote accepted

$C = \bigcap\limits_{i\in I} (A_i \cup T_i)$, 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 $0$.)

share|cite|improve this answer
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.