# Construct a set of real numbers whose limit points comprise the set of integers $\mathbb{Z}$

My thought process is the following: Let $S=\{ m + \frac{1}{n}| m \in \mathbb{Z},n \in N \}$. Then I need to show that the limit points of $S$ are indeed the integers and that these are the only limit points. I don't know where to go from here.

• How would you prove that $2$ is a limit point? – user208259 Jan 22 '15 at 22:00
• What was your thought process when you came up with $S$ in the first place? – sciona Jan 22 '15 at 22:01
• @user208259 Let $m=2$ and $n$ run through all the naturals? I am so confused about the definitions about metric spaces. I am knew to analysis and have read this section in Rudin like 10 times and still and befuddled haha.. – Jack Jan 23 '15 at 1:17
• @sciona my thought process was that if we fix $m$ and then run through all the naturals, we will attain $m$ as the limit point. Then if we do this for all $m \in \mathbb{Z}$ we get back limit points which are exactly the integers. – Jack Jan 23 '15 at 1:19
• @Tim Using Rudin's notation, you want to check that every neighbourhood $N_r(2)$ contains a point in $S$. How can you show that there is a point in $S$, other than $2$, that is at a distance of less than $r$ from $2$? You're on the right track. – user208259 Jan 23 '15 at 1:58

Your example is correct because the $\lim_{n\to \infty} m+\frac{1}{n} = m\in \mathbb{Z}$ for all $m\in \mathbb{Z}$, however your trick here is that you use that $m\in\mathbb{Z}$.

• Yeah I was thinking that as well. I am just confused if I need to add anything more than that. Also, The question says to construct a set of real numbers, but I used the integers. I know that the integers are in the reals but I do not know if the professor explicitly meant use all the reals. – Jack Jan 23 '15 at 1:10
• It seems that when he goes away from the questions in Rudin and makes up his own, they are sometimes ambiguous. This happened on a previous assignment. – Jack Jan 23 '15 at 1:11
• It's not like you'd be able to use all the reals anyway (your example only uses rationals, for example). – Squirtle Jan 23 '15 at 1:21
• That is true. I was just wondering if there is a way to use all the reals to get limit points which are only the integers. Sorry if I am slow to the draw here, this is my first analysis class and I am trying to grasp all the new definitions still.. – Jack Jan 23 '15 at 1:24
• Well the cardinality of the reals is more than the integers, so for every integer there are many, many sequences you can use from the reals (which ever: irrationals, etc) – Squirtle Jan 23 '15 at 1:36

Your example works, and it actually does not contain any integers if you check it closely. Another trivial example would be the integers themselves. But for your set, it's not hard to show that any integer $m \in \mathbb{Z}$ is a limit point:

Take any arbitrary neighborhood $N_r(m)$ around $m$ (meaning that $r$ is an arbitrary, positive real number). By definition, this will be of the form $N_r(m) = \{s \in S : |s - m| < r \}$. Now find an element of $S$ that is inside of this set. Thus any neighborhood of $m$ contains an element of $S$, so $m$ is a limit point.

To show that any other point is not a limit point, fix a noninteger point $m + \frac{1}{n} \in S$. Now find some real number $r > 0$ such that $N_r(m + \frac{1}{n})$ contains no other points of $S$. Thus some neighborhood of any noninteger point contains no elements of $S$, so no noninteger point is a limit point.

For example, take the number $5 + \frac{1}{10}$. The closest point in $S$ to this number is $5 + \frac{1}{11}$, and the distance between them is $\frac{1}{10} - \frac{1}{11}$. So any neighborhood around $5 + \frac{1}{10}$ of radius $r = \frac{1}{10} - \frac{1}{11}$ or less will not include this point. You can actually take the radius to be $r = \frac{1}{10} - \frac{1}{11}$ because a neighborhood is defined to be all points of distance strictly less than $r$, thus it will not include a point that is exactly distance $r$ away.

• This statement is not true: Another trivial example would be the integers themselves. – Dave L. Renfro Jan 23 '15 at 19:53
• I'm sorry, you're right. The integers have no limit points. – KidA424 Jan 23 '15 at 21:37
• "and it actually does not contain any integers" eeerr what about when $n=1$ – mercio Jan 23 '15 at 21:50
• Again, good point! If you don't want it to contain any integers, you could use $n \in \mathbb{N} - \{1\}$ instead of $n \in \mathbb{N}$. – KidA424 Jan 24 '15 at 8:43