# show that the finite intersection of open sets is non empty. for the cocountable Topology.

Let $$X$$ be an uncountable set and let $$\tau = \lbrace \phi \rbrace \cup \lbrace U \subset X : U^c$$ is countable $$\rbrace$$.

1) i) Prove the countable intersection of open sets in $$\tau$$ remains open.

ii) Is this result true in usual $$\mathbb{R}$$?

• i) Let $$A_i \in \tau$$, how can I write the countable intersection ? is it like that $$(\bigcap A_i)^c$$ ?

If it is correct, then

$$(\bigcap A_i)^c =\bigcup A_i^c$$

The countable union of countable set is a countable set, and then it closed, thus it's complement which is $$\bigcap A_i$$ open set.

• ii) NO, $$\cap (\frac{-1}{n},\frac{1}{n})=\lbrace 0 \rbrace$$.

2) Show that the finite intersection of open sets is non empty. Is this result true in usual $$\mathbb{R}$$?

Let $$A, B\in \tau$$, and suppose that $$A \cap B=\phi$$, then $$A \subset B^c$$, since $$B \in \tau$$, then $$B^c$$ is countable, and hence $$A$$ also is countable, But it is not necessarily to be that, so our supposition is not correct.

3) Is $$\overline{\mathbb{Q}} = \mathbb{R}$$? for the topology $$\tau$$. what about $$\mathbb{Q}^c$$ and $$[0,1]$$.

No, $$\mathbb{Q}$$ is closed, since it is countable, then $$\overline{\mathbb{Q}}=\mathbb{Q}$$.

but what about for $$\overline{\mathbb{Q}^c}$$?, the closure is the smallest closed set which is contains $$\mathbb{Q}^c$$, in this topology, it will be the smallest countable set which is contains $$\mathbb{Q}^c$$, and this is impossible, since $$\mathbb{Q}^c$$ is not countable, the same thing with $$[0,1]$$.

Is that true, please?

"sorry I don't speak English well".

The essence of the argument you gave is OK.

Suppose that $$U_n$$, $$n \in \mathbb{N}$$ are open sets in the co-countable topology. If one of them is $$\emptyset$$, then the intersection is $$\emptyset$$ too, hence open, so we can assume that the $$U_n$$ all have countable complement.

Then $$(\bigcap_{n \in \mathbb{N}} U_n)^\complement = \bigcup_{n \in \mathbb{N}} U_n^\complement$$ by de Morgan, so the complement of $$\bigcap_{n \in \mathbb{N}} U_n$$ is a countable union of countable sets, hence countable. So $$\bigcap_{n \in \mathbb{N}} U_n$$ is open.

Your example for the usual topology is fine. Maybe prove that $$\{0\}$$ is not open, as you claim?

b) If $$U$$ and $$V$$ are both non-empty open (so countable complements) and suppose $$U \cap V = \emptyset$$, then indeed $$U^\complement \cup V^\complement = \mathbb{R}$$ and the reals would be countable, contradiction.

In the usual topology $$(0,1)$$ and $$(2,3)$$ are disjoint non-empty open sets.

c) Indeed $$\mathbb{Q}$$ is closed, being countable, and so equals its own closure in this topology.

$$[0,1]$$ and $$\mathbb{Q}^\complement$$ are dense: the only closed sets are the countable sets and $$\mathbb{R}$$ so any uncountable set has only one containing closed set.