Rationals as a dense subset of the reals.

I am having difficulty with a problem in one of my textbooks. It gives three definitions

$1$. A set is called a closed set if its complement is open.

$2$.The closure of a set E is the intersection of all Closed sets C of which E is a subset.

$3$.If is a sets closure is equal to $\mathbb R^n$, then it is dense.

The question asks to show that the set of rational numbers are a dense subset of the reals. This is a second year university textbook. Here is what I have so far: To show that the of set rational numbers is dense, it suffices to show that its closure is the set of Real Numbers. The closure of the set of rational numbers is defined to be the intersection of closed sets containing the rational numbers, and is thus the smallest closed set containing the rational numbers. A real number is defined to be the limit of a sequence of rational numbers, and therefore the smallest set that can contain all rational numbers (of which there exists a rational number between every two real numbers) as well as their boundary points (the irrational numbers) is the set of real numbers.

• What you have is correct, but if the intent is to prove it from the 3 definitions given, there is no discussion of boundary points in them, so your proof relies concepts that go beyond this definition. I think what you need to prove directly from those definitions is the result that every non-empty open interval contains a rational number. From this it follows that all non-empty open sets contain a rational, and therefore by (1), every closed set other than $\Bbb R$ cannot contain every rational number. – Paul Sinclair Sep 28 '15 at 3:42
• Thank you. I am also somewhat confused why the set of Real numbers is a closed set.. – IntegrateThis Sep 28 '15 at 3:46
• @JamesDickens what is your definition of an open set? Why does the empty set fit this definition? – Omnomnomnom Sep 28 '15 at 3:58
• A set is open if and only if it contains an open interval about each of its elements. Since the empty set has no elements, it satisfies this condition vacuously, so it is open. It's complement is $\Bbb R$, so $\Bbb R$ is closed. Note that the empty set and $\Bbb R$ are both open and closed. Those are the only two subsets of $\Bbb R$ with this property. – Paul Sinclair Sep 28 '15 at 4:01
• Yes this is the definition of open that I have. As a side question what is the complement of the rationals. Also why is the complement of the empty set the Real numbers and not the complex numbers or even that the complement of the Real numbers is also the complex numbers. I appreciate the help.. I am clearly somewhat confused. I understand that the empty is closed by the vacuous reasoning though. – IntegrateThis Sep 28 '15 at 4:08

Let's say the closure of $\mathbb Q$ is a set ($\mathbb Q \subsetneq) X$ which is a proper subset of $\mathbb R$ ($X\subsetneq \mathbb R$).

Our aim is to show $X$ cannot be closed i.e. to show $\mathbb R -X$ is not open.

For any point $a\in \mathbb R- X$ for any open set $U$ containing a there exists a rational number in the open set $U$. Therefore any open set $U\not \subset R-X$ ( because $\mathbb R -X$ doesn't contain any rational point (as $\mathbb Q \subset X$)).

So, $\mathbb R -X$ is not open, and thus $X$ is not closed.

Therefore, any proper subset $X$ of $\mathbb R$ is not closed. So the smallest closed subset of $R$ containing $\mathbb Q$ is nothing but $\mathbb R$.

Conclusion- $\mathbb Q$ is dense in $\mathbb R$

Aside: About your question why the complement of empty set is $\mathbb R$ not $\mathbb C$.

When you talk of complement you always have an ambient space or an universal space to start off. So you will see the answer if you re-frame your question. Instead of questioning, "what is the complement of empty set?" one should question "what is the complement of empty set in $\mathbb R$. And the answer should be obviously $\mathbb R$.

If one asks "what is the complement of empty set in $\mathbb C$, then the answer will be $\mathbb C$.

I wish this helps.

• We say a set $X$ is closed in $\mathbb{R}$ when its complement is an open set in $\mathbb{R}$. You want to show that $X$ cannot be closed (then it has to be open), therefore its complement $\mathbb{R} - X$ must be closed. But you state that "Our aim is to show $X$ cannot be closed i.e. to show $\mathbb{R}-X$ is open." How come? – DrHAL Apr 27 '16 at 1:45
• @DrHal Thank you for noticing, that was a typo, it should be $\mathbb R -X$ is not open. I have corrected it. – Babai Apr 27 '16 at 6:01
• @DrHAL On the other hand I don't agree to the part of your comment where you say "You want to show that X cannot be closed (then it has to be open)," $X$ cannot be closed doesn't mean $X$ has to be open. A subset can be neither open nor closed. What we can say is the if $X$ is not closed then the complement is not open. – Babai Apr 27 '16 at 6:05