Understanding the definition of a limit point.

Question

I am trying to understand the following definition of a limit point:

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

I am trying to understand what "other than $x$" means. I know that a limit point $x$ need not be in $A$. So, suppose we know somehow that $x \notin A$. Then, would it be valid to modify the definition of a limit point as follows?

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

EDIT: I found a different definition of a limit point:

A point $x$ is a limit point of a set $A$ iff $x = \lim a_n$ for some $(a_n) \subseteq A$ satisfying $a_n \neq x$ $\forall n \in \mathbb{N}$

Does this definition fail when it comes to the sequence $(a, a, \dots)$ whose limit point clearly is $a$?

Answer 1

Let $\mathbb{R}$ be given the usual metric.

The idea of limit points of a subset $A \subset \mathbb{R}$ is to say that in the neighbourhood of such points, I can find at least one other point different from that point which is also in $A$.

So if $x \notin A$, then $x$ is a limit point of $A$ iff in any $\varepsilon$-neighbourhood of $x$, I can find at least one point of $A$. This is precisely your modified definition. So yes, it should be valid.

Addendum:

For the edited part: Note the condition $a_n \neq x$ for all $n \in \mathbb{N}$ carefully. This condition precisely prevents the constant sequence that you mentioned.

If this condition is not present, then every point in a set $A$ could be a limit point of $A$. This is not necessarily true. Consider the following set \begin{equation} S = \{ 1/n : n \in \mathbb{N}\}. \end{equation} Then none of the points in $S$ is a limit point of $S$. Moreover, the only limit point of $S$ is $0$ which is not in $S$.

Finally, the definition does not fail with the constant sequence as it is sufficient to find just one such sequence that satisfies $x = \lim a_n$.

Answer 2

Yes, if $x$ is not in $A$, then the definition you gave is equivalent. I'll give an example of where it does matter:

We can define $A=\left[0,1\right]\cup\left\{ 2\right\} $. In this example, $2$ isn't a limit point even though it is in the set. This is because we can choose $\left(1.5,2.5\right)$, which is an open set containing only $2$ (which is in $A$) and no other points of $A$.

Answer 3

Both the definitions are equivalent and imply that $x$ is limit point of set A if every $\epsilon\gt 0 $ neiborhood of $x$ contains infinitely many points of $A$ (other than $x$).
For if, on the contrary some $\epsilon$ neiborhood of $x$ contains finitely many points of A. Without loss of generality, assume that the neibourhood of $x$ contains only $m \gt 0$ points of A viz. $a_1,a_2,...,a_m$ then define $\delta = min\{|x-a_i|: i=1,2,3,...,m\}$ .
Now take $\epsilon \lt \delta$ and consider $\epsilon$ neiborhood of $x$. This neiborhood does not contain any point of $A$, which is a contradiction as $ x$ was supposed to be limit point of set $A$.
In your question, what is set $A$? It is a finite set $\{a\}$ because the sequence is a constant sequence $(a,a,a,....)$. Hence it does not have a limit point.

Answer 4

"I am trying to understand what "other than x" " What else could it mean? If $x \in A$ but it is a singleton point and there is a neighborhood around $x$ that contains no other point of $A$ except $x$ then that is not a limit point. Even though every neighborhood does contain $x$ itself. And $x \in A$.

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Consider the the set $\mathbb Z\subset \mathbb R$ and the "usual" metric.

Is $1.999$ a limit point of $\mathbb Z$?

If you take any $\epsilon$ no matter how small, then the $\epsilon$-neighborhood around $2$ is $(1.999-\epsilon, 1.999+\epsilon)$. And if $\epsilon \le 0.001$ then $there is no integer in that neighborhood.

So $1.999$ is not a limit point.

Now does $\mathbb Z$ have any limit points?

Before we answer that, let's ask, SHOULD $\mathbb Z$ have any limit points?

$\mathbb Z$ is a collect of distinct points separated from each other. It "shouldn't" have any limit points points, because a limit point should be a point the no matter how close you get to it, there's going to be a point in the set right there close to it. And with $\mathbb Z$ if we start taking epsilons less than $1$ then all the points in t set are all going to be isolated and alone. If you start getting closer than one to any point, that points should be isolate and apart from all integers.

So what about $2$? Is $2$ a limit point of $\mathbb Z$. Well, if you take any $\epsilon: 0 < \epsilon < 1$ then the $\epsilon$-neighborhood $(2-\epsilon, 2 + \epsilon)$ doesn't have any integers ....

... except $2$.... Evey $\epsilon$-neighborhood of $2$ has $2$ itelf in it.

So be your definition, $2$ is a limit point.

But.... it shouldn't be. To say "there is always an integer close to $2$" shouldn't count for $2$ itself. It's.... weird ... to say that $2$ is "close to" $2$. This misses the point of limit points entirely.

A limit point should be that every neighborhood of $x$, no matter how small, you can find a element of $A$. But $x$ itself doesn't count. You need to find an element of $A$ that is "close to" $x$; not the point $x$ itself.

If $x \not \in A$ that's not an issue. Everypoint of $A$ that is close to $x$ won't be equal to $x$ because $x$ isn't in $A$. But if $x\in A$ then to be a limit point the every $\epsilon$-neighborhood of $x$ will contain a point of $A$ (other that $x$ itself).

That's really all it means.

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2). That alternative definition fails if $\lim a_n = x$.

Consider $\mathbb Z$ which has no limit points. ANd $x = 1.999$ and $y=2$.

Let $a_i \in \mathbb Z$. And let $\lim a_i = 1.999$. That is clearly impossible. SO there is no sequence of integers hows limit is $1.999$, so $1.999$ is not a limit point. That makes sense. You can't "get close" to $1.999$ because you will always be at least $0.001$ away.

Now le $a_i \in \mathbb Z$ and $a_i\ne 2$. Then $\lim a_n = 2$ is imposibble. You can't "get close" to $2$ because you will always be $1$ away. So $2$ can't be a limit point.

But what if $a_1 = 1, a_2 =3$ and for $i > 2; a_i =2$.. Then $\lim a_n = 2$. But that doesn't count. You aren't "getting close to $2$". You are jumping across a brook and landing smack dab on $2$. That's not "getting close". That's stomping around with you big boots on and being there.

Answer 5

The right word IMHO is accumulation point or cluster point instead of usual limit point as these words give you the meaning directly without too much symbolism.

A point $x$ is said to be an accumulation point of a set $A$ if an infinite number of points of $A$ are accumulated or clustered near $x$. Formally this is defined by requiring that every neighborhood of $x$ must have some point of $A$ other than $x$.

The requirement of "other than $x$" is because if this is dropped then every member of $x\in A$ is a limit point as trivially every neighborhood of $x$ contains $x$ which is a member of $A$ also. The idea of infinitely many points accumulating near $x$ goes for a toss. In fact by relaxing this requirement you see that finite sets also have limit points. And you guess that its fishy as analysis almost always deals with infinite.

And yes the definition in later part of your question is equivalent to the standard definition. Since there are infinitely many points of $A$ clustered near $x$ it is not a big deal to construct a sequence from these which tends to $x$. You can write a formal proof by taking neighborhoods of the form $(x-1/n,x+1/n)$ and find an $x_n\in A$ which differs from $x$ and lies in that neighborhood.

The notion of a limit point of a sequence is a bit different. If $\{x_n\} $ is a sequence then a number $L$ is said to be a limit point of the sequence $\{x_n\} $ is for every $\epsilon >0$ and every positive integer $m$ there is a positive integer $n$ such that $n>m$ and $x_n\in(L-\epsilon, L+\epsilon) $. Under this definition a constant sequence also has a limit point. And in general if $A$ denotes the range of the sequence (set of values taken by the sequence) then limit point of set $A$ and limit point of the sequence are not same.