# Thread: Show no point is a limit point of M

1. ## Show no point is a limit point of M

I hope this is the right forum, I wasn't sure exactly where to post this. The course is Analysis I. I wrote a proof for this problem, but I am new at these and I'm unsure whether my logic holds. All we are allowed to use for this assignment is the definition of limit point. Here is what I have.

Problem: Show that if $M$ is the set of all positive integers, then no point is a limit point of $M$.

Solution: Assume that $\exists$ a limit point $p$ of $M$. Then all open intervals containing $p$ must contain an element of $M$ different from $p$. Consider the interval $I = (s, q)$ where $s \in M$and $q \in M$ and $s < p < q$. That means $\exists$ a point $r \in M$ such that $s < r < q$. But if $q = s+1$, then no such $r$ exists because any point between $s$ and $s+1$ is not an integer. Thus, not every open interval containing $p$ contains a point of $M$ different from $p$, therefore no limit point of $M$ exists.

My question is mainly in regards to the case where I suggest $q = s+1$ because does that also mean I'm implying that $p \notin M$ which isn't necessarily true? Or is this okay?

2. ## Re: Show no point is a limit point of M

you have two cases for the p, if p is a positive integer you can take the open interval (p-1/2,p+1/2 ) which dose not contains any other element of M except p
if p is not in M, p located between two positive integers say n, n+1 now let "t" be the minimum of |p - n| , |p - (n+1)| so the interval (p-t , p+t) contains p and intersect with M at p just
if p is less than 1 the interval (- infinity , 1 ) contains p and dose not intersect with M.

Solution: Assume that $\exists$ a limit point p of M. Then all open intervals containing p must contain an element of M different from p. Consider the interval I = (s, q) where s \in M and q \in M and s < p < q. That means \exists a point $r \in M$ such that s < r < q. But if q = s+1, then no such r exists because any point between s and s+1 is not an integer. Thus, not every open interval containing p contains a point of M different from p, therefore no limit point of M exists.
why you consider just the open intervals with positive integers as a bound ?

3. ## Re: Show no point is a limit point of M

The first thing to ask is, what is the ambient space in which you're being asked about limit points - AND what is that space's topology? That's unstated in your problem. Perhaps M is itself the encompassing space? Perhaps with the topology inherited from the reals? Or, more likely, the set M is being considered as a subset of $\mathbb{R}^1$ with it's usual topology? There are limitless possibilities. When you wrote "Consider the interval $I = (s, q)$ where $s \in M$and $q \in M$ and $s < p < q$", you choose an unexpected open set, making me wonder about exactly what the ambient space, and its topology, is.

But I'll assume that the problem is "Show $M = \mathbb{Z} \subset \mathbb{R}^1$ (with its usual topology) has no limit points." I suspect you just went astray in your choice of interval. There's no need to have the endpoints of the interval be in $M$. What you're trying to show is that there's an open set OF $\mathbb{R}^1$(!!) that contains $p$ and yet has no intersection with $M$ - well except for $p$ if $p \in M$. But the open set needs only to be open in the surrounding space.

Here are two examples that I think might clarify the situation:

Ex1: $p = 4.0387$ is not a limit point of $M = \mathbb{Z} \subset \mathbb{R}^1$ (with its usual topology).
Proof: Let $U$ be the interval $(4.03, 4.04)$. Then $p \in U$ and $U$ open in $\mathbb{R}^1$.
But since $U \cap M = \emptyset$, it follows that $p$ cannot be a limit point of $M$ (since otherwise the intersection would contain something other than $p$).

(Note that IF I had chosen $U = (3.5, 4.5)$, then $U \cap M = \{4\}$, then, since $4 \ne p$, I couldn't use that interval to conclude that $p$ isn't a limit point of $M$. We know $p$ isn't a limit point - in fact I just proved it - but IF I had poorly chosen the interval, that interval wouldn't have served to prove that it isn't.)

Ex2: $p = 12$ is not a limit point of $M = \mathbb{Z} \subset \mathbb{R}^1$ (with its usual topology).
Proof: Let $U$ be the interval $(11.5, 12.2)$. Then $p \in U$ and $U$ open in $\mathbb{R}^1$.
But since $U \cap M = \{12\} = \{p\}$, it follows that $p$ cannot be a limit point of $M$ (since otherwise the intersection would contain something other than $p$).

Now - try to prove this for any generic $p \in \mathbb{R}^1$.

4. ## Re: Show no point is a limit point of M

Thank you for the help, the examples make sense. So I have thought this through a little more and I get what you are saying, now I am just trying to word it properly. Tell me how this sounds:

Solution: Assume that $\exists$ a limit point $p$ of $M$. Then all open intervals containing $p$ must contain an element of $M$ different from $p$. Consider an interval $I$ that contains $p$. If $p \in M$, then any such $I$ bounded by $(p-1, p+1)$ does not contain any other points of $M$ different from $p$ because no other integers lie in this range. If $p \notin M$, then any $I$ containing $p$ that is bounded by $(a, a+1)$ where $a \in M$ will not contain any elements of $M$ because no integers lie in this range. Thus, not every open interval containing $p$ contains a point of $M$ different from $p$, therefore no limit point of $M$ exists.

5. ## Re: Show no point is a limit point of M

Originally Posted by Chaobunny
Thank you for the help, the examples make sense. So I have thought this through a little more and I get what you are saying, now I am just trying to word it properly. Tell me how this sounds:
Solution: Assume that $\exists$ a limit point $p$ of $M$. Then all open intervals containing $p$ must contain an element of $M$ different from $p$. Consider an interval $I$ that contains $p$. If $p \in M$, then any such $I$ bounded by $(p-1, p+1)$ does not contain any other points of $M$ different from $p$ because no other integers lie in this range. If $p \notin M$, then any $I$ containing $p$ that is bounded by $(a, a+1)$ where $a \in M$ will not contain any elements of $M$ because no integers lie in this range. Thus, not every open interval containing $p$ contains a point of $M$ different from $p$, therefore no limit point of $M$ exists.
To show that $p$ is not a limit point of $M$ all that is necessary is to show one open set containing $p$ and no other point of $M$.

6. ## Re: Show no point is a limit point of M

Originally Posted by Plato
To show that $p$ is not a limit point of $M$ all that is necessary is to show one open set containing $p$ and no other point of $M$.
It was not necessary for me to show an example for both cases of $p$, where $p \in M$ and $p \notin M$? I thought that was what the above users were trying to tell me. Is there a more general way to say this without splitting it into two cases?

7. ## Re: Show no point is a limit point of M

Originally Posted by Chaobunny
It was not necessary for me to show an example for both cases of $p$, where $p \in M$ and $p \notin M$? I thought that was what the above users were trying to tell me. Is there a more general way to say this without splitting it into two cases?
That might be the easiest way. But my point was that all one has to do is find one open set containing $p$ and no other point of $M$. Again, using two cases may be easiest for you.

8. ## Re: Show no point is a limit point of M

Ah, okay, thank you!

9. ## Re: Show no point is a limit point of M

Your 1st post (the one where you begain this thread) was a bit confusing to read. Your 2nd post was much easy to follow, and was correct.

However, I might criticize one part if it:
"If $p \notin M$, then any $I$ containing $p$ that is bounded by $(a, a+1)$ where $a \in M$ will not contain any elements of $M$ because no integers lie in this range."

Here's how I might do it:

Choose any p in R.
If p in M, let U = (p-0.5, p+0.5) be an open set of R containing p. Then U intersect M = {p}. Thus p isn't a limit point of M.
If p not in M, then there's an integer n s.t. n<p<n+1. Let U = ((n+p)/2, ((n+1)+p)/2).
Note that the left endpoint of U is the midpoint between n & p, and the right endpoint of U is midpoint between (n+1) & p.
Note that p in U. (Proof: n<p implies that n+p<2p, implies that (n+p)/2 <p. Also p<n+1 implies 2p < n+1+p implies p < (n+1+p)/2.)
Then p cannot be a limit point of M, because U intersect M is obviously empty.
(But just to prove that: Assume U intersect M is not empty. Let m be in the intersection. Then m in M (i.e. m an integer), and m in U.
Since m in U, have that (n+p)/2 < m < ((n+1)+p)/2.
Also, n<p implies that 2n<n+p, which implies that n<(n+p)/2.
Likewise, p<n+1 implies n+1+p<2(n+1) implies (n+1+p)/2 < n+1.
Since (n+p)/2 < m < ((n+1)+p)/2, it follows that n < (n+p)/2 < m < ((n+1)+p)/2 <n+1, so n<m<n+1. But that's impossible for integers n and m.
Thus the assumption that U intersect M is not empty was false. Therefore U intersect M is empty.)