# Why is $S=\{U \subseteq \mathbb R:\mathbb R \setminus U \text{ is finite}\}$ a topology?

I have a worksheet for my real analysis course that sets up a problem by defining "a new topology $$S$$ on $$\mathbb R$$ by declaring a set $$U \subseteq \mathbb R$$ to be in $$S$$ if $$\mathbb R \setminus U$$ is finite." We haven't covered topologies in class yet, but the worksheet says that one of the conditions for a thing $$T$$ to be a topology is that $$\emptyset \in T$$.

It seems to me that $$\mathbb R \setminus \emptyset$$ would not be finite, because it would simply be $$\mathbb R$$ which is not finite. So $$\emptyset \subset S$$, but $$\emptyset \notin S$$. This would mean $$S$$ is not a topology. Have I misunderstood something about topologies?

• The cofinite topology is defined to contain all cofinite subsets, as well as the empty set. – Jean-Claude Arbaut Oct 16 '19 at 23:22
• You are correct, you have to explicitly add the empty set to $S$ to make it a topology. This is the so-called “co-finite” topology. – Thomas Andrews Oct 16 '19 at 23:22
• So as literally written, S would not be a topology? If so, is it a convention to implicitly add the empty set as needed when defining a topology? – Jacob Jones Oct 16 '19 at 23:24
• Now that I've got a name, it looks like math.stackexchange.com/q/1876571/698711 is related. – Jacob Jones Oct 16 '19 at 23:28

You're correct, $$\mathbb{R} \setminus \emptyset = \mathbb{R}$$ which is not finite. To be a topology the empty set must be included and $$S$$ doesn't include the empty set. You seem to understand it correctly.