Hi Folks,

Can anyone help with Q. 52 GR058. For reference, the question is:

Q. If "A" is a subset of the real line "R", and "A" contains each rational number, which of the following must be true?

A. If "A" is open, then "A" = "R"

B. If "A" is closed, then "A" = "R"

C. If "A" is uncountable, then "A" = "R"

D. If "A" is uncountable, then "A" is open

E. If "A" is countable, then "A" is closed.

Thanks,

## Q.52 GR0568

A. NO. Counterexample, A=(-inf,sqrt(2)) union (sqrt(2),+inf).

B. YES. But leave it for a while.

C. NO. Counterexample, A=Q union [0,1].

D. NO. Previous counterexample also works here. In this case A is neither open, nor closed.

E. NO. Counterexample A=Q, which is countable but not closed.

Therefore we got the answer.

Notice that we could get it immediately, if remembered that closed set contains all its limit points. For Q the set of its limit points is R.

- my understanding was that since the set "A" contains each rational number, it must at least be equal to Q or more. Can we deduce from the wording of the question that "A" may contain elements that are not in Q? Can you give a counterexample to option A. for a set which contains all elements in Q (and maybe more).

- Secondly, could you explain choice B.

Thanks again.

Amateur.

Correct.my understanding was that since the set "A" contains each rational number, it must at least be equal to Q or more.

Correct.Can we deduce from the wording of the question that "A" may contain elements that are not in Q?

I just did it in the previous post, didn't I?Can you give a counterexample to option A. for a set which contains all elements in Q (and maybe more).

The set of all limit points of Q is R. In order to have all its limit points A must be R.Secondly, could you explain choice B.

### Re: Q.52 GR0568

Sorry to pull up this old thread, but I'm just slightly confused about the answer. I understand why all the incorrect choices are incorrect, but I did not think that R was closed.

### Re: Q.52 GR0568

It has to be! The empty set is always open, so its complement must be closed.jayre wrote:Sorry to pull up this old thread, but I'm just slightly confused about the answer. I understand why all the incorrect choices are incorrect, but I did not think that R was closed.

Additionally, R must contain all its limit points, as it's the whole space.