Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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 was skimming through the topology book recommended through this question and I came across a question that I apparently solved incorrectly. It's question 2.5 on page 12.

Repeated here:

Let $X$ be $\mathbb{R}$, and let $\Omega$ consists [sic] of the empty set and complements of all finite subsets of $\mathbb{R}$. Is $\Omega$ a topological structure?

And my answer is "no", because I can posit some finite intersection of infinite sets and come up with a finite set, which is no longer in $\Omega$. For example, for some $a,b \in \mathbb{R}$ and $a<b$, two infinite sets are $X_1 = (-\infty,b)$ and $X_2 = (a,\infty)$ and obviously, $X_1, X_2 \in \Omega$ but $X_1 \cap X_2 \notin \Omega$.

But it seems it is a topology, because in the line following the question, they state:

The space of Problem 2.5 is denoted by $\mathbb{R}_{T_1}$ and called the line with $T_1$-topology.

So where am I going wrong?

share|cite|improve this question
Your $X_1$ and $X_2$ are not the complements of finite sets, so they're not in $\Omega$. – Chris Eagle Mar 17 '11 at 23:19
the finite complement topology is often used as an example of a non-hausdorf topology, as any two non-empty open sets intersect. – yoyo Mar 17 '11 at 23:38
up vote 3 down vote accepted

The problem with your argument is that not all infinite subsets of $\mathbb R$ are complements of finite sets. To see that we in fact have a topology, let $X_1$ and $X_2$ be finite subsets of $\mathbb R$ and observe that $(\mathbb R - X_1) \cap (\mathbb R - X_2) = \mathbb R - (X_1 \cup X_2)$ which is again the complement of a finite set. The union property should be obvious.

share|cite|improve this answer
As I've been told. But your equation is quite helpful. Thanks! – JasonMond Mar 17 '11 at 23:34

Not all infinite subsets of $\mathbb{R}$ are complements of finite subsets of $\mathbb{R}$. For example, $X_1=(-\infty,b)$ is the complement of $[b,\infty)$, which is infinite. Thus $X_1$ is not open in this topology.

share|cite|improve this answer
Oh, crud. You're right. Thanks! – JasonMond Mar 17 '11 at 23:22

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.