# Confusing definition of limit points

I found this definition on wikipedia.

A point x in X is a limit point of S if every neighbourhood of x contains at least one point of S different from x itself.

But doesn't this just mean it could be pretty much any point? Not necessarily anywhere near a boundary/limit of S? Just a random point in S.

For example pick the point, 2, in S (the standard topology of R between (1,3)). Every neighbourhood of x will contain a point in S that is not x.

What am I missing, this notion of a limit point seems pointless..?

• EVERY nbhd of x. Mar 20, 2016 at 12:22
• Yes, because 2 IS a limit point. but take the set $\mathbb Z$. It doesn't have any limit points. Feb 4, 2017 at 19:15
• All interior points are (counterintuitively) limit points but not all limit points are interior points. All boundary points are limit point but no boundary point is an interior point. Isolated points are neither interior or limit. All interior points and isolated points are in the set. Limit and boundary points may or may not be. So the definition is not trivial or useless. Feb 4, 2017 at 19:48

Intuitively, when we say that some $x \in X$ is a limit point of $S$, we mean that that there are infinitely many elements of $S$ that are densely packed near $x$. Indeed, we can get arbitrarily close to $x$ using only elements of $S$.

For example, you may have heard that $\mathbb Q$ is dense in $\mathbb R$. A consequence of this fact is that any irrational number, such as $x = \pi$, is a limit point of $S = \mathbb Q$ and thus can be approximated to an arbitrary precision using only rational numbers such as: $$3, 3.1, 3.14, 3.141, 3.1415, 3.14159, \ldots$$

The word limit in the definition of limit point its not related to the concept of topological boundary. It is related to the fact that it is the limit of a sequence of points in $S$ different from $x$.

Also, not every point of $S$ is a limit point. Isolated points like $2$ in $(0,1)\cup\{2\}$ are not limit points.

• cool thanks. I think I get it now, I was just considering continuous open sets for which limit point don't help much, but there are plenty of other uses in discontinuous sets. Mar 20, 2016 at 8:18
• Actually limit points are of consideration in continuous open sets as the interior points are limit points but the boundary points are too. But the boundaries are not in the set. So for an open set with boundaries there are limit points not in the set so the set is not closed. (In general... there are exceptions.) Feb 4, 2017 at 19:43

In your example the set $(1,3)$ is open, so every point is a limit point - try to proof this for a arbitrary open set for better understanding. But now, for example, have a look at the set $(0,1] \cup \{2\} \cup [4,8)$. Here is not every point a limit point - try to find them. So this notion is not pointless only because there are some sets in which every point is a limit point.

• haha yea, now you mention it, it does seem a little silly to say "limit points are pointless only because there are some sets in which every point is a limit point." Mar 20, 2016 at 8:17

Everyone has already written nice answers. I just wanted to show some pictures to build an idea bout the definition . If you see for $(0,1)$ take any neighbourhood of $0$ (as it says every nbhd of $x$ in the definition of the limit point) you will always find a point of $(0,1)$  For the second one if you see you can find a sufficiently large $\epsilon$ such that the intersection between $[0,1]$ and $B_{\epsilon}(2)$ will be nonempty but not for every neighbourhood.

You are confusing limit point with boundary points.

In $S= (1,3)$ the point $2$ is a limit point. But the point $0$ or $27$ are not. Every point in $[1,3]$ is a limit point.

So yes, every point of $S$ is a limit point. That is not a problem.

But take the set $S = \mathbb Z$. No point of $S$ is a limit point as for any $n \in S$ then neighborhood $(n-1,n+1)$ contains no point of $S$ other than $n$.

Think of it this way. A limit point is not a point that is "at the limit" of $S$. A limit point is a point that is "within limits" of $S$. Supposed you lived in Hobokon, New Jersey. Then you are right on the city limits of New York City. But supposed your friend Bill lived deep in the heart of Manhattan. Then Bill lives within the New York city limits. So both you and Bill are in limit points of New York City.

(Thus New York City is an open set... we can extend this further. Tijuana, Mexico; and Topeka, Kansas are limit points of the United States. But the U.S. embassy in Paris, France, though while considered part of the United States is not a limit point because no neighborhood under a hundred miles [I'm pretty sure] will contain any point of the United States that is not the embassy.)

(The United states is neither closed nor open. Australia is closed as there are no limit point outside Australia. If Australia were to close its foreign consulates and embassies, then Australia would be both closed and open.)

(But I digress....)

But doesn't this just mean it could be pretty much any point?

Yes, of course. But if the space is a Hausdorff space (almost all interesting spaces are), after you've selected such a point, I just select another, smaller neighbourhood that doesn't contain that point you chose (in a Hausdorff space I always can do that), and you will have to give me a point in that neighbourhood as well. Indeed, I can repeat that process infinitely often, and since each time you'll be forced to present me yet another point, the end result is that there will have to be infinitely many points in any neighbourhood.

Not necessarily anywhere near a boundary/limit of S?

A limit point doesn't need to be near a boundary. Indeed, for $\mathbb R$ with the usual topology, every point is a limit point of the full set $\mathbb R$, and that set doesn't even have a boundary (or more exactly, its boundary is the empty set)!

Now on the distance of the chosen point from the limit point, of course if you even want to speak of "near the limit", you have to have a concept of nearness; usually in the form of a metric. And in that case, I can play essentially the same game as above, except this time taking the distance into account:

I start by providing some neighbourhood, and you give me some point in the neighbourhood. Given that we are now in a metric space, that point will have a certain positive distance from the limit point. Thus I present you in the next step the neighbourhood that only consists of those points that are closer than half that distance. So now you'll have to provide me with another point in that neighbourhood, which means especially a point that is less than half as far away from the limit point as the point you've originally chosen. Again, I can repeat that infinitely often, and unless you run out of points to give me (in which case the point wasn't a limit point to begin with), you'll end up with a sequence of points that come arbitrary close to the limit point (the sequence of distances clearly converges to zero).

On the other hand, if you don't have such nice properties like Hausdorff or even metric, then you might indeed get away with choosing just one point. For example, take any set with the trivial topology. Then whatever point you chose, there's only one neighbourhood: The complete space. So indeed, in that space, all you need for a point to be a limit point of a set is to have one other point that belongs to the set.