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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".

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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.

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