10

I need to show that there exist sequences s.t. for fix $\epsilon>0$ there exist $|z_n-\alpha|<\epsilon$ (1) holds for infinitely many $n\in N$ but s.t. $\alpha$ is not a limit point of the set containing all terms $z_n$.

Thus far I've basically constructed a sequence with several limit points but that does not converge and I guess that's way to go. What confuses me that if I have infinitely many $n \in N$ that satisfy (1) how can there be an epsilon neighborhood around $\alpha$ which contains no points of $z_n$? Is this the archimedean principle at work? My hunch is that im first choosing $N(\epsilon)$ and then in then in then in the set part choosing $\epsilon(N)$.

I'd love any answer that gets me any closer to understanding this and/or the archimedean principle at work here.

Thanks

/I

Winston
  • 864

1 Answers1

11

Let $x_n=(-1)^n$ and $\alpha=1$. Then $S=\{x_n:n\in\Bbb N\}=\{-1,1\}$ is a finite set, so it has no limit points.

The point of this exercise is that cluster points of sequences and limit points of sets aren’t quite the same thing. A point $x$ is a cluster point of the sequence $\sigma=\langle x_n:n\in\Bbb N\rangle$ if some subsequence of $\sigma$ converges to $x$; $x$ is a limit point of the set $S=\{x_n:n\in\Bbb N\}$ if every open nbhd of $x$ contains a point of $S\setminus\{x\}$. If $x$ is a limit point of the set $S$, then every open nbhd of $x$ contains infinitely many different points of $S$; if $x$ is a cluster point of $\sigma$, then every open nbhd of $x$ contains infinitely many different terms of $\sigma$, but those terms might all be the same point, as in the example that I gave above.

Brian M. Scott
  • 616,228
  • Brian, can I ask a question related to this topics? – Frank Swanton Oct 20 '16 at 01:53
  • 1
    @Frank: Sure. Is it something simple that you can ask here, or does it need a new question? – Brian M. Scott Oct 20 '16 at 08:51
  • Thanks, Brian. Well, I wanted to post as a new question, but I was afraid that the admin would direct me to this question, because it is related. But I of course want to credit any answer or response from you, so I can post a new question, but here is some of the confusion I am having. – Frank Swanton Oct 20 '16 at 12:09
  • I learned from intro to analysis that limit point of a sequence is a point where the sequence visits infinitely many times. So in the example, it would ping pong back to -1 and 1, so both are limit points. But I also read from a book, the set $S$, you defined is the image of the sequence, which is precisely what you have defined in your answer. Basically here, I see that we are making a leap from a sequence to a set of #s. So I understand this, because you can think of a sequence as a function from naturals to the space, say reals. – Frank Swanton Oct 20 '16 at 12:12
  • My confusion is the limit point on this set $S$, image you defined. The way I read from some texts is the following: $L$ is a limit point of a set if you have a sequence converging to it. Another definition I read is, if every open ball around it contains at least one point of A distinct from itself, which is the definition you wrote in your answer. So, in this definition, we cannot use the sequence of singletons -1, -1, -1, -1, ..., correct? Because this would be just itself. So what if we had a set $S=[0,1]\cup{5}$? – Frank Swanton Oct 20 '16 at 12:15
  • In the case of $S$ I created, any point in the closed interval [0,1] would be a limit point of $S$, correct? Because you can produce a convergent sequence in the closed interval. 5 would fail to be a limit point, because there exists an open ball where the intersection between that open ball and $A${5} is empty. Does this mean, accumulation point of set and limit point of set is also different? Just FYI, I am studying metric space, the usual $(X,d)$. I had a very good grip on these concepts on real line, but I got confused as soon as image of sequence came in and introducing a set. thanks! – Frank Swanton Oct 20 '16 at 12:19
  • 1
    @Frank: The first definition of limit point of a set works for metric spaces (and, more generally, for first countable spaces) if you require that the sequence be in $S\setminus{x}$, but it’s wrong for spaces in general; the second definition is the correct one. A constant sequence has a cluster point (to which it converges, so it’s actually the limit of the sequence), but the corresponding set has only one point — in your example it’s the set ${-1}$ — and therefore has no limit point. The limit points of $[0,1]\cup{5}$ are the points of $[0,1]$; $5$ is not a limit point of that set. – Brian M. Scott Oct 20 '16 at 12:20
  • 3
    @Frank: If $S$ is a set in a topological space, the statements ‘$x$ is a limit point of $S$’, ‘$x$ is an accumulation point of $S$’, and ‘$x$ is a cluster point of $S$’ all mean exactly the same thing: that every nbhd of $x$ contains a point of $S\setminus{x}$. In metric spaces this is equivalent to saying that there is a sequence in $S\setminus{x}$ that converges to $x$. – Brian M. Scott Oct 20 '16 at 12:23
  • Brian, thank you for the response. This really brings some clarity. So is my understanding of the following then correct? Suppose we have a sequence of 5,5,5,5,... This sequence has a limit point 5. However, the image of the sequence which is S={5} does not have a limit point, because every open ball u create around 5, the intersection between 5 and the set minus 5 which is empty is always empty. But we know S is closed. So it must contain all of its limit points. It doesn't have any. So vacuously it is closed? – Frank Swanton Oct 20 '16 at 12:45
  • 1
    @Frank: Yes, a set with no limit points is automatically closed. Some examples in $\Bbb R$ are any finite set, $\Bbb N$, and $\Bbb Z$. And yes, the constant sequence at $5$ has $5$ as its limit, but the set ${5}$ has no limit point. The sequence defined by $x_n=(-1)^n$ has cluster points $-1$ and $1$, but the set ${-1,1}$ has no limit points. (I prefer to use the term cluster point for points that are limits of subsequences of a sequence and reserve limit point for the point (if any) to which the sequence converges.) – Brian M. Scott Oct 20 '16 at 12:49