# Conflicting definition of limit point

This question was raised at a different place without sufficient answers.

Definition 1: We say that a vector $x \in R^n$ is a limit point of a sequence $\{x_k\}$ in $R^n$ if there exists a subsequence of $\{x_k\}$ that converges to $x$. This definition appears frequently in the optimization literature, for instance, see Bertsekas, Nonlinear Programming, 2nd edition, page 666.

But a definition of limit point in real analysis is different.

Definition 2: A point $z_0$ is a limit point for a set of point if every neighborhood of $z_0$ contains points, other than $z_0$ of set.

Accordingly, $a_n=\{5, 4, 3, 2, 1, 0, 0, ...,\}$ has a limit point of $0$ based on the first definition, but $0$ is not a limit point based on the second definition.

The problem is that in the literature, limit point is frequently used but without a definition. So what is the default definition in computer science or optimization? Is Definition 1 the default? For instance, what is the presumed definition of limit point in https://arxiv.org/pdf/1209.2385?

• If you're talking about sequences, clearly Definition 1 (which is the standard definition of a limit point of sequences in real analysis) is meant, and Definition 2 (which talks about sets) is irrelevant. Of course, there's a relation: To every sequence $\{a_n\}_{n\in\mathbb{N}}$, you can associate the set $\{a_n:n\in\mathbb{N}\}$. Then, every limit point of the sequence is a limit point of the set, and every limit point of the set is a limit point of the sequence or recurs infinitely often. (Constant sequences are somewhat degenerate cases.) – Christian Clason May 5 '18 at 23:15

The definition of the limit of a sequence is really rather uncontroversial and can be found at https://en.wikipedia.org/wiki/Limit_of_a_sequence . Intuitively, a sequence $\{a_1,a_2,\ldots\}$ converges to a limit $a$ if the $a_k$ come closer and closer to $a$. Sitting on top of $a$ is allowed. So $\{4, 3, 2,1, 0,0,0,0,\ldots\}$ converges to zero just fine.

The concepts you are confused by are as follows:

• In compact sets, every sequence has one or more subsequences that converges to something. For example, the sequence $\{1,0,1,0,1,0,\ldots\}$ does not converge, but it has subsequences that converge to zero or one. This seems to be what you suggest in your definition 2 (though I don't think that it is quite correct as stated).

• In the paper you reference, the sequence is indexed by something else. For example, let there be a sequence $\{a_1(x),a_2(x),a_3(x),\ldots\}$ for each choice of $x$. Then depending on $x$, these sequences may converge to different limits $a(x)$. Statements such as those in the preprint you reference say that for any choice of $x$, the limit point $a(x)=\lim_{k\rightarrow \infty} a_k(x)$ has certain properties. (In the example of the preprint, $x$ is the starting point and the statement says that no matter where you start, you will always converge to a stationary point. Which stationary point you converge to may be different, though!)

• For example, the sequence {1,0,1,0,1,0,…} does not converge, but it has subsequences that converge to zero or one. I thought this is based on definition 1, but not definition 2. Now for the referenced paper, what is the definition of limit point in your argument "the limit point $a(x)=lim_{k\to\infty} a_k(x)$ has certain properties." – John Smith May 4 '18 at 19:41
• @JohnSmith : That is the point-wise limit of a sequence of functions, plain old convergence. – Lutz Lehmann May 4 '18 at 20:52
• @LutzL: But, on page 7 in the proof of the refereed paper, it says: "Assume that there exists a subsequence $\{x^{r_j}\}$ converging to a limit point z." This seems to suggest utilizing definition 1, not the "plain old convergence." – John Smith May 4 '18 at 21:12
• @JohnSmith : But that is also a different situation than the formula in your last comment. Function sequence $a_k(x)$, point sequence $x^r$, $k,t\in\Bbb N$. And yes, they mean the ordinary limit $\lim_{j\to\infty}x^{r_j}$, the limit for the fixed subsequence that was chosen so that it converges. This usually uses compactness of level sets. – Lutz Lehmann May 4 '18 at 22:31
• @Lutzl: I don't think there is a function sequence issue like $a_k(x)$. Here $x^r, r \in \Bbb N$ is a point or vector sequence in the paper. If $z$ is a limit point of the $x^r$, it is a stationary point of some problem as defined in the paper. Where does $a_k(x)$ come from is beyond my understanding. – John Smith May 4 '18 at 23:51