Closure of a countable subset of the reals

[radou]

Homework Statement

Let A = {ln(1 + q^2) : q is rational}. One needs to find Cl(A) in R with its euclidean topology.

The Attempt at a Solution

So, the set A is a countable subset of [0, +∞>. The closure is, by definition, the intersection of all closed sets containing A. So, Cl(A) would be [0, +∞> itself , right?

I'm just curious about my solution, thanks in advance.

By the way, another way to look at it would be the fact that A is dense in [0, +∞> (since every open interval in [0, +∞> intersects A), so Cl(A) = [0, +∞>. I'm not really sure about this, although it seems quite obvious. Can we, for every <a, b> in [0, +∞>, find some rational q such that ln(1 + q^2) is contained in <a, b>?

 

[Eynstone]

As the rationals are dense in R, the closure will be the range of f(x) =ln(1 + x^2) i.e.,
[0, +∞).

 

[radou]

As the rationals are dense in R, the closure will be the range of f(x) =ln(1 + x^2) i.e.,
[0, +∞).

OK, thanks. But I was not sure about that, since the values of f are not rational in general.

 

[HallsofIvy]

Yes, that's why the range is all real numbers larger than or equal to 0 and not just rational numbers.