Is there an intuitive, not-too-mathematical way of thinking about limit points?

Question

so I know this question has been asked sooo many times. But I just have a few questions in particular, which despite searching, I haven't found an answer to. I appreciate any help.

Book's definition:

A point $x$ is a limit point of a set $A$ if every $\epsilon$-neighborhood $V_{\epsilon}(x)$ of $x$ intersects the set $A$ in some point other than $x$.

Particularly, wanted clarification with the bolded part of this sentence.

If we take our set $A = [1,4), A \in \mathbb{R}$, then would the point $\{2\}$ be a limit point of the set $A$? Since every epsilon neighborhood of $2$ intersects $A$ in a point other than $2$.

Or does that bolded sentence mean to imply that, the limit point itself, cannot belong to the set $A$? So in this case, that would mean to imply, only $4$ is a limit point of the set $A$ ?

Is it possible for someone to give a diagrammatic example of what is and isn't a limit point?

Edit: I am trying to get a "non-mathematical" idea of a limit point, and thus I will opt to not mark this question as a duplicate.

I mean, technically one could envision and understand this definition with the $\epsilon$ definition, but for people like me who are mathematically challenged, it's nice to have a different way of thinking about it (they are the exact same idea, I know, but this helps me to link and understand the concept). MY question has been solved, now, however.

Answer 1

$2$ is a limit point, as is $4$. Your set can contain its limit points, in fact, in metric spaces, a closed set is defined to be one that contains all of its limit points.

Here's a picture that might help http://upload.wikimedia.org/wikipedia/commons/thumb/b/ba/Interior_illustration.svg/220px-Interior_illustration.svg.png

Answer 2

$2$ is a limit point of $[1,4)$ because every neighborhood of $2$ contains some other point of $[1,4)$. It doesn't matter that $2$ is in the set (or not).

For example, if you take the $\epsilon$ neighborhood $V_{\epsilon}(2)$ as you mention above, then if $\epsilon\geq 1$, you can use that $2.5\in V_{\epsilon}(2)\cap[1,4)$ and $2.5\not=2$. On the other hand, when $\epsilon<1$, you can use $2+\epsilon/2\in V_{\epsilon}(2)\cap[1,4)$ and $2\not=2+\epsilon/2$ since $\epsilon>0$.

Answer 3

$2$ would be a limit point of $A$. In fact all points of $[1,4]$ are limit points of $A$. The bolded sentence means exactly what it says, nothing more nothing less. A limit point $x$ is a point if every neighborhood contains a point of $A$ other then $x$, or that $(V_\epsilon(x)\setminus \{x\})\cap A$ is nonempty.

Answer 4

The first statement is the correct one.

In other words, we say that $x$ is a limit point of $A$ if $(V_\epsilon(x) \cap A) - \{x\} \neq \emptyset$ for all $\epsilon$.

Answer 5

I find limit points to be easier to understand when you throw out the metric baggage.

A point $x$ is a limit point of $A$ if every neighborhood of $x$ overlaps with $A-{x}$.

Or, a point $x$ is a limit point of $A$ if every neighborhood of $x$ overlaps with some point of $A$ that isn't $x$.

We subtract ${x}$ even if $x$ is not in $A$ because every neighborhood of $x$ already includes $x$, so if we didn't the test would be useless.

Now all we need to do is talk about neighborhoods. Many different definitions work, from epsilon-balls, to any set containing an open set. The basic idea is that everything that $x$ "embedded inside" of is a neighborhood of $x$. How we decide "embedded inside" can vary.

In a metric space (a space with a metric, aka a concept of "distance" that acts distance-like), we can take our neighborhoods to be all open balls of radius epsilon>0. Or any set that contains such a ball around the point if we prefer.