No! An arbitrary union of C-open sets is not C-open. It follows a counterexample in R, with usual topology:

Let A1 be the union of intervals (-1,1), (1,2),...,(p,p+1),(p+1,p+2),...

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For natural k>1, Ak= the union of (-k,k), (k,k+1),...,(p,p+1),(p+1,p+2),...

cl(Ak)\Ak={-k,k,k+1,...,p,p+1,p+2,...}=the union of N\{0,1,..,k-1} and {-k}, which is countable.

The union of all Ak when k natural >0 is the whole R, and cl(R)\R is the empty set!

But if we define that a set is C-open if cl(A) \ A is at most countable(i.e. countable, finite or the empty set), then the above counterexample do not works!

Dear Dinu,

Thank you so much for your wonderful response

A set S is countable if there exists a subset of the natural numbers, say T, and a one-to-one correspondence f between S and T. Since emptyset is a subset of the natural number, then it is countable

Thank you very much for your great answer. However, look at the definition of the countable complement topology on the real numbers R. The set R is open because its complement is equal to phi which means phi is countable.

Dear Mesfer,

I use the following definition for countable set:

A set S is countable if there exists a one-to-one correspondence f between S and N. (N being the set of natural numbers).

It's true the fact that many books give the definition you posted above. But I consider wrong to work with phi and finite sets as countable sets! This can lead to ambiguities regarding cardinality.

Dear Dinu,

Thank you so much.