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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.

Math1000
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Jack
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  • How would you prove that $2$ is a limit point? – user208259 Jan 22 '15 at 22:00
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    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 01:17
  • @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 01:19
  • @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 01:58
  • Is it because the rationals are dense in the reals? So that if $r= \frac{1}{n}$, then by denseness, there exists a rational $p$ such that it is between 2 and $\frac{1}{n}$. So the distance from 2 to $p$ is less than the distance 2 to $\frac{1}{n}$. Therefore every neighborhood of 2, contains a point $p\neq 2$ in $S$. Thus 2 is a limit point. – Jack Jan 23 '15 at 03:54
  • No, that's not right. It's not the case that such a rational number $p$ must belong to $S$. The distance between $2$ and $2 + 1/n \in S$ is $1/n$. So what you must show is that, no matter how small $r$ is, there is some $n$ for which $1/n < r$. That inequality is equivalent to $n > 1/r$. So you can pick any integer $n$ greater than $1/r$. Now you must extend the same argument to any integer $m$, as well as proving that any number that is not an integer cannot be a limit point. – user208259 Jan 23 '15 at 04:01
  • @user208259 Thanks for your help! I was going to retort but I am still kind of confused. I hope my professor can cover this in lecture tomorrow because he hasn't touched this yet. I will work more on the problem this weekend and will post when I get somewhere. Thank you so much for your time! I really appreciate it! – Jack Jan 23 '15 at 06:09

2 Answers2

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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 $m\in\mathbb{Z}$.

Squirtle
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  • 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 01:10
  • 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 01:11
  • 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 01:21
  • 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 01:24
  • 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 01:36
  • For example, consider Warren Hill's answer here:http://math.stackexchange.com/questions/849274/the-convergence-of-sqrt-2-sqrt-2-sqrt-2-ldots – Squirtle Jan 23 '15 at 01:41
  • Even more: in the example above even though the limit is an integer (rational) every element in that sequence is irrational; suppose by way of contradiction that a rational element were in this sequence, square it (still rational), then subtract two (still rational), what's left is the previous term in the sequence, we can repeat this process until we get to the first term in the sequence, $\sqrt{2}$, but this shows that is rational, which is a contradiction. Therefore every term in that sequence is irrational even though the limit is rational. Corollary: the irrationals are not closed. – Squirtle Jan 23 '15 at 03:48
  • I see what your saying, that the limit of a sequence or set, does not necessarily need to be in said set or sequence. Is that correct? I guess I am just getting confused over the term limit point and its definition. – Jack Jan 23 '15 at 04:00
  • Limit points are the set of points which are limits of sequences where those sequences are in the set. Every element $x$ in a given set is a limit point of that set because the constant sequence ($a_i=x$ for every $i$) has limit $x$, so $x$ (any point of said set) is a limit point. My example above shows not all limit points are from a given set (I.e the irrationals have $2$ as a limit point). Indeed all reals can be found by shifting this sequence by the desired amount (term by term, ie if every element in sequence has 1.5 added to it then the limit will be 3.5, in this way you can get all R – Squirtle Jan 23 '15 at 04:10
  • thank you that makes sense! – Jack Jan 23 '15 at 06:14
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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.

Michael Knopf
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