# Limit points and interior points

I am reading Rudin's book on real analysis and am stuck on a few definitions.

First, here is the definition of a limit/interior point (not word to word from Rudin) but these definitions are worded from me (an undergrad student) so please correct me if they are not rigorous. The context here is basic topology and these are metric sets with the distance function as the metric.

A point $p$ of a set $E$ is a limit point if every neighborhood of $p$ contains a point $q \neq p$ such that $q \in E$

Also, an interior point is defined as

A point $p$ of a set $E$ is an interior point if there is a neighborhood $N_r\{p\}$ that is contained in $E$ (ie, is a subset of E).

I understand interior points. Ofcourse given a point $p$ you can have any radius $r$ that makes this neighborhood fit into the set. Thats how I see it, thats how I picture it.

I can't understand limit points. It seems trivial to me that lets say you have a point $p$. Then one of its neighborhood is exactly the set in which it is contained, right? ie, you can pick a radius big enough that the neighborhood fits in the set.

Ofcourse I know this is false. Our professor gave us an example of a subset being the integers. He said this subset has no limit points, but I can't see how.

• "Then one of its neighborhood is exactly the set in which it is contained, right? ie, you can pick a radius big enough that the neighborhood fits in the set." -- I don't understand what you are saying clearly, but this seems wrong. The correct statement would be: "No matter how small an open neighborhood of $p$ we choose, it always intersects the set nontrivially." – Srivatsan Feb 1 '12 at 4:23
• For the integers, you can take any $n \in \mathbf Z$ and $N_r(n)$ for $r \leq 1$, and this will show that $n$ is not a limit point. – Dylan Moreland Feb 1 '12 at 4:29
• I ran into the same problem as you, I made a question a few months ago (now illustrated with figures)! math.stackexchange.com/questions/93288/… – Samuel Reid Feb 1 '12 at 10:04

I understand in your comment above to Jonas' answer that you would like these things to be broken down into simpler terms.

Think about limit points visually. Suppose you have a point $p$ that is a limit point of a set $E$. What does this mean? In plain terms (sans quantifiers) this means no matter what ball you draw about $p$, that ball will always contain a point of $E$ different from $p.$

For example, look at Jonas' first example above. What you should do wherever you are now is draw the number line, the point $0$, and then points of the set that Jonas described above. Namely draw $1, 1/2, 1/3,$ etc (of course it would not be possible to draw all of them!!).

Now an open ball in the metric space $\mathbb{R}$ with the usual Euclidean metric is just an open interval of the form $(-a,a)$ where $a\in \mathbb{R}$. Now we claim that $0$ is a limit point. How?

Given me an open interval about $0$. For now let it be $(-0.5343, 0.5343)$, a random interval I plucked out of the air. The question now is does this interval contain a point $p$ of the set $\{\frac{1}{n}\}_{n=1}^{\infty}$ different from $0$? Well sure, because by the archimedean property of the reals given any $\epsilon > 0$, we can find $n \in N$ such that

$$0 < \frac{1}{n} < \epsilon.$$

In fact you should be able to see from this immediately that whether or not I picked the open interval $(-0.5343,0.5343)$, $(-\sqrt{2},\sqrt{2})$ or any open interval.

Now let us look at the set $\mathbb{Z}$ as a subset of the reals. What you do now is get a paper, draw the number line and draw some dots on there to represent the integers. Can you see why you are able to draw a ball around an integer that does not contain any other integer?

Having understood this, looks at the following definition below:

$\textbf{Definition:}$ Let $E \subset X$ a metric space. We say that $p$ is a limit point of $E$ if for all $\epsilon > 0$, $B_{\epsilon} (p)$ contains a point of $E$ different from $p$.

$\textbf{The negation:}$ A point $p$ is not a limit point of $E$ if there exists some $\epsilon > 0$ such that $B_{\epsilon} (p)$ contains no point of $E$ different from $p$.

From the negation above, can you see now why every point of $\mathbb{Z}$ satisfies the negation? You already know that you are able to draw a ball around an integer that does not contain any other integer.

• Hey just a follow up question. In Rudin's book they say that $\mathbb{Z}$ is NOT an open set. But how can this be? If I draw the number line, then given any integer I can draw a ball around it so that it contains two other integers. And this suffices the definition for an interior point since we need to show that only ONE neighbourhood exists. – Tyler Hilton Feb 1 '12 at 23:15
• @Tyler Write down word for word here exactly what the definition of an interior point is for me please. – user38268 Feb 1 '12 at 23:33
• A point p is an interior point of E if there is a nbd $N$ of p such that N is a subset of E. – Tyler Hilton Feb 1 '12 at 23:44
• @TylerHilton More precisely: A point $p$ of a subset $E$ of a metric space $X$ is said to be an interior point of $E$ if there exists $\epsilon > 0$ such that $B_\epsilon (p)$ $\textbf{is completely contained in }$ $E$. Now when you draw those balls that contain two other integers, what else do they contain? They also contain reals, rationals no? So how is the ball completely contained in the integers? – user38268 Feb 1 '12 at 23:50
• okay got it! So is this the reason why $E=\{\frac{1}{n}|n=1,2,3\}$ is not closed and not open? its not closed well because 0 is a limit point of it (because of the archimedan property). Why is it not open? I can pick any point $p=\frac{1}{n}$ and choose an interval so that the nbd is contained in E. From your definition this would fail because this interval also includes reals? – Tyler Hilton Feb 2 '12 at 0:02

Consider the set $\{0\}\cup\{\frac{1}{n}\}_{n \in \mathbb{N}}$ as a subset of the real line. Let's consider 2 different points in this set. First, let's consider the point $1$. Is it a limit point?

No. In fact, if we choose a ball of radius less than $\frac{1}{2}$, then no other point will be contained in it. So it's not a limit point.

Consider the point $0$. For any radius ball, there is a point $\frac{1}{n}$ less than that radius (Archimedean principle and all). So for every neighborhood of that point, it contains other points in that set. Thus it is a limit point.

Does that make sense?

• Yes! thankyou. if you didnt mention the fact that there was an intersection with the set that contained zero, it would still have 0 as as intersection point, right? – Tyler Hilton Feb 2 '12 at 0:10
• It was helpful that you mentioned the radius. – user66360 Oct 15 '13 at 23:53
• This comment hepled me a lot! Thank-you – elbarto Feb 22 '15 at 7:24

The definition of limit point is not quite correct, because $p$ need not be in $E$ to be a limit point of $E$.

Note that for $p$ to be a limit point of $E$, every neighborhood of $p$, no matter how small, must intersect $E$ in points other than $p$. So if there is a small enough ball at $p$ so that it misses $E$ entirely (unless $p$ happens to be in $E$), then $p$ is not a limit point. When $p$ is a limit point, there are points from $E$ arbitrarily close to $p$.

Examples:

• In $\mathbb R$, $0$ is a limit point of $\left\{\frac{1}{n}:n\in\mathbb Z^{>0}\right\}$, but $-1$ is not.
• In $\mathbb R$, $\mathbb Z$ has no limit points. For each $p\in\mathbb R$, there is a closest integer $n\neq p$, and the ball of radius $|p-n|$ centered at $p$ does not intersect $\mathbb Z$ (except perhaps at $p$).

If $p$ is a not a limit point of $E$ and $p\in E$, then $p$ is called an isolated point of $E$. This is good terminology, because $p$ is "isolated" from the rest of $E$ by some sufficiently small neighborhood (whereas limit points always have fellow neighbors from $E$). If $p$ is not in $E$, then not being a limit point of $E$ is equivalent to being in the interior of the complement of $E$.

• Thankyou. I understand that a little bit better. Would it be possible to even break it down in easier terms, maybe an example? I am having trouble visualizing it (maybe visualizing is not the way to go about it?) – Tyler Hilton Feb 1 '12 at 4:35
• Sorry Tyler, I've done all I can for now. Best wishes, – Jonas Meyer Feb 1 '12 at 4:37

For a limit point $p$ of $E$ (where $p$ does not need to be in $E$ to start with, so that part of the definition is wrong) we need that every neighbourhood of $p$ intersects $E$ in a point different from $p$.

Let's see why the integers $\mathbb{Z} \subset \mathbb{R}$ do not have limit points: if $x$ is not an integer then let $n$ be the largest integer that is smaller than $x$, then $x$ is in the interval $(n, n+1)$ and this is a neighbourhood of $x$ that misses $\mathbb{Z}$ entirely, so $x$ is not a limit point of $\mathbb{Z}$.

Of course there are neighbourhoods of $x$ that do contain points of $\mathbb{Z}$, but this is irrelevant: we need all neighbourhoods of $x$ to contain such points. So to show a point is not a limit point, one well chosen neighbourhood suffices and to show it is we need to consider all neighbourhoods.

Continuing the proof: if $x = n$ is some integer, then $(n-1, n+1)$ is a neighbourhood of $x = n$ that intersects $\mathbb{Z}$ only in $x$, so this again shows that $x$ is not a limit point of $\mathbb{Z}$: one neighbourhood suffices to show this, again.

For a positive example: consider $A = (0,1)$. Then every point of $A$ is a limit point of $A$, and also $0$ and $1$ are limit points of $A$ that are not in $A$ itself. To see this for $0$, e.g., any neighbourhood $O$ of $0$ contains a set of the form $(-r,r)$ for some $r > 0$, and then $r/2$ is a point from A, unequal to $0$ in $(-r,r) \subset O$, and as we have shown this for every neighbourhood $O$, $0$ is a limit point of $A$.

i was reading this post trying to understand the rudins book and figurate out a simple way to understand this. In a limit point you can choose ANY distance and you'll have a point q included in E, on the other hand in an interior point you only need ONE distance so that q is included in E

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