closure of (0,1) on R with the co-countable topology

Question

So I am learning topology on my own from a script. Earlier it defines what a co-finite topology is. Now I am trying to answer the question: "Consider R with the co-countable topology. Show that the closure of (0,1) is the whole space. On the other hand show there is no sequence in (0,1) converging to 2."

Now by analogy I would assume the co-countable topology is the collection of sets such that their complement is countable (and the empty set). However, I am very confused: since the complement of (0,1) in R is most certainly not countable, (0,1) should be closed. Therefore the closure of (0,1) would be (0,1), not R. However when I looked around on the internet a bit, it seemed like the closure is indeed R. What am I missing? Thank you in advance, I'm sorry for asking what I'm sure is a "dumb" question.

Answer 1

You misunderstood the concept of closed set. A set is closed if its complement is open.

In the case of co-countable topology, a set $A$ is closed if its complement is open, i.e. if $A^c=\mathbb{R}-A$ is open. That is if $$(\mathbb{R}-A)^c=\mathbb{R}-(\mathbb{R}-A)$$ is countable, which means $A$ itself is countable.

So, yes, the complement of $(0,1)$ in $\mathbb{R}$ is not countable, so $(0,1)$ is not open but it doesn't mean it is closed. We may have sets that are neither open nor closed.

Answer 2

With respect to this topology, the open subsets of $\mathbb R$ (other than $\emptyset$ and $\mathbb R$) are those sets $A$ such that $A^\complement$ is countable (in this context, finite sets are countable too). In other words, a subset of $\mathbb R$ (other than $\emptyset$ and $\mathbb R$) is closed if and only if it is countable.

Therefore, the only closed subset of $\mathbb R$ of which $(0,1)$ is a subset is $\mathbb R$ itself. In other words, $\overline{(0,1)}=\mathbb R$.