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I always thought they were the same thing. The limit of a sequence is a point such that every neighborhood around it contains infinitely many terms of the sequence. The limit point of a set is a point such that every neighborhood around it contains infinitely many points of the set. So is the limit of a sequence also a limit point of a set that contains it? what's the difference?

user42
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3 Answers3

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Let's work in the real line, for concreteness. Consider a sequence $(x_n)_{n \geq 1}$. A limit point of the sequence is a limit point of the set $\{x_n \mid n\geq 1\}$. When you say the limit point, it means that the set $\{x_n \mid n\geq 1\}$ has only one limit point, say, $L$. And this $L$ is the element who satisfies the definition we all know and love: $$\forall \ \epsilon > 0, \ \exists \ n_0 \geq 1, {\rm s.t.} \ n > n_0 \implies |x_n - L| < \epsilon.$$

If the set $\{x_n \mid n\geq 1\}$ has more than one limit point, then the sequence $(x_n)_{n \geq 1}$ does not converge.

Ivo Terek
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  • I guess I didn't ask the question very well. What I meant is,if pn is a sequence of some subset A of a metric space X, is the limit of pn (denoted p) also a limit point of A? – user42 Dec 17 '14 at 02:39
  • basically is the limit of a sequence also a limit point of the set that it is a part of – user42 Dec 17 '14 at 02:40
  • No. Take a metric space with the discrete metric. The only convergent sequences are constant ones (from some $n$ onward). Then, for example, take $X = {1,2}$ with the discrete metric. Consider $A = {2}$ and $p_n = 2$ for all $n$. Then $p = 2 \in A$, but it's not a limit point of $A$. – Ivo Terek Dec 17 '14 at 02:43
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    What you can always say is: $p_n \in A \ \forall \ n \implies p \in \overline{A}$ (closure). – Ivo Terek Dec 17 '14 at 02:44
  • oh okay. thanks so much! – user42 Dec 17 '14 at 02:57
  • @Ivo Terek, your example works because with the discrete metric any point is isolated, specificcaly the point 2 is isolated and on the other hand your sequence $p_n$ is constant. I think that the idea of user1961722 is correct, it just need some additional condition, I mean if $p_n$ converges to p and $p_n$ is in $A$ then $p$ is a limit point of $A$ provided that the sequence $p_n$ is not constant! – Idris Addou Dec 17 '14 at 04:44
  • I could take the sequence $(1,2, 1,2,1,2,2, 2, 2, 2,\ldots)$ and have the same conclusion. Not eventually constant. But yes, I agree, with some extra conditions it should be true – Ivo Terek Dec 17 '14 at 04:53
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    A limit point of a sequence $(x_n)_{n\geqslant1},$ convergent or not, need not be a limit point of the set ${x_n\colon n\geqslant1}.$ What is true without exception is that a limit point of a sequence is a point that either occurs infinitely often in the sequence or is a limit point of the set of points in the sequence. Or so I recently argued here, but my answer to that question seems to have gone down like a lead balloon, so perhaps I slipped up? (Excuse the late comment, but this question has just been brought to the front page by a new answer.) – Calum Gilhooley Apr 22 '20 at 14:23
  • I don't want to downvote the answer. I'd much rather the error was corrected. Do you dispute the accuracy of the comment I left six weeks ago? I'll happily delete it (and this one) if it's mistaken. – Calum Gilhooley Jun 05 '20 at 20:34
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    A limit point of a sequence does not necessarily mean it is the limit point of the range set. e.g. $ (-1)^n$ has limit points $1, -1$ but the range set $ {1,-1}$ has no limit points. The converse however, is always true. – Faqir Chand Nov 18 '21 at 18:04
  • A sequence and the set of values of the sequence are not the same! @FariqChand's comment can be taken as an example. – Emo Jan 29 '22 at 12:59
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In a general metric space X, it depends on the range of the sequence whether both the notions are same or not. Let $x_n$ be a convergent sequence i.e. its limit exists and is unique, say $x$. Then its range is a set, say E $\subseteq$ X, which contains all the distinct elements in the sequence $x_n$.

If the set E is finite, then it is obvious that one of the terms in E, suppose $y$, occurs infinitely many times in the sequence $x_n$. Then the limit of the sequence is, of course, $y$. Also, the $y$ is unique since the sequence is convergent so every subsequence has same subsequential limit. So, $y = x$.
But note that E has no limit point since it is finite (a point is a limit point of a subset of a metric space if all of its neighbourhoods contain infinitely many points of the subset) and in particular, all of its points are isolated points.

Now if the range set E is countably infinite (it can't be uncountable since the sequence itself is countable), then E has a unique limit point equal to $x$. Hence, every neighbourhood has infinitely many points of E and hence of $x_n$. Then for a given $\varepsilon \gt 0,$ $$\exists N \in \Bbb N \ni d(x_n,x)\lt \varepsilon\quad \forall\;n \ge N$$ Limit point is unique because if it was not the case, then there would be another point say $z\, (\neq x)$ in $\overline{E}$ which has the above property. Hence the limit of the sequence would not be unique.
Thus, if the sequence $x_n$ has infinitely distinct elements, then both the notions are same.

P.S: If you consider a subset A such that A $\supseteq$ E, then as commented by littleO, A can have many limit points, different from the limit of the sequence, no matter whether E is finite or infinite. But for this case, the question of differentiating between the two notions also becomes irrelevant.

Singh_Gunjeet
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Let's take elements of a sequence as a set say S. S has many subsets, which form many subsequences of the primary sequence. S has a limit that would be the limit of sequence & some of it's subsets too have limits those would be the limits of their respective subsequences & the limit point of whole sequence Can say that the set of all limits of subsets would be limit points of the sequence.

Mathematics
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