Understanding the definition of cluster point in real sequence

Question

I'm reading the definition of cluster point in Royden's Real Analysis. It is as follows.

$\ell$ is a cluster point of $<x_n>$ if, given $\epsilon>0$ and given $N$, $\exists n \geq N$ such that $|x_n- \ell| < \epsilon$.

Prior to reading this definition, my understanding of cluster point $\ell$ was that there should be INFINITELY MANY points of the sequence in ANY neighborhood of $\ell$. However, with the definition given in Royden's book, I don't see any conditions that enforce the existence of infinitely many points of a sequence and any neighborhood? In fact, I think one should reword the definition as follows.

$\ell$ is a cluster point of $<x_n>$ if, for any $\epsilon>0$, there exists infinitely many $x_n$ such that $|x_n- \ell| < \epsilon$.

Is there anything wrong with my rationale?

Answer 1

The definition does secretly enforce the existence of infinitely many $x_n$. Notice that $N$ is given. Thus, if you find some $x_k$ such that $|x_k - l| < \varepsilon$, then you can again apply the definition using $N' > k$. You can then find $x_{k+a}$ where $|x_{k+a} - l| < \varepsilon$.

Answer 2

Both definitions you give are in fact equivalent. Lets denote them as

(1) For any $\def\eps{\varepsilon}\eps > 0$ and any $N \in \mathbf N$, there exists $n \ge N$ such that $\def\abs#1{\left|#1\right|}\abs{x_n - \ell}<\eps$.
(2) For any $\eps > 0$ there exist infinitely many $n \in \mathbf N$ such that $\abs{x_n - \ell} < \eps$.

Suppose (1) holds, to prove (2) let $\eps > 0$ be given. We inductively will find a striclty increasing sequence $(n_k)$ in $\mathbf N$ such that $\abs{x_{n_k} - \ell }< \eps$. Choose $n_1 \ge 1$ according to (1) with $N=1$. If $n_k$ has been constructed, use (1) with $N = n_k + 1$, giving an $n_{k+1}\ge n_k + 1 > n_k$ such that $\abs{x_{n_{k+1}} - \ell} < \eps$. As $\{n_k : k \in \mathbf N\}$ is an infinite subset of $\mathbf N$, we have proven (2).

Suppose (2) holds, let $\eps > 0$ and $N \in \mathbf N$ be given. According to (2), choose an infinite subset $A \subseteq \mathbf N$ such that $\abs{x_n - \ell}< \eps$, $n \in A$. As $A$ is infinite, we cannot have $A \subseteq \{1, \ldots, N\}$, hence there is sone $n \in A$ with $n \ge N$. So we have found an $n \ge N$ with $\abs{x_n - \ell}<\eps$.

Answer 3

The first definition implies that there are infinitely many $n$ so that $|x_n - L| < \varepsilon$ holds.

Let's assume otherwise, i.e. the set $M = \{ n \mid x_n - L| < \varepsilon\}$ is finite. Since $M$ is finite, the set admits a maximum $N$. But by our assumption, we can find an $n_0 \ge N + 1$ to that $|x_{n_0} - L| < \varepsilon$ holds. But this is a contradiction, since $n_0 > \max M$ implies that $n_0$ is not in $M$.

Another way to see this: For every $n_k$ with $|x_{n_k} - L| < \varepsilon$ we can find a $n_{k + 1} \ge n_k + 1$ so that $|x_{n_{k + 1}} - L| < \varepsilon$ holds. This construction yields infintely many $n$ that satisfy $|x_n - L| < \varepsilon$.

Answer 4

The first definition says that the set of $n$ such that $|x_n- \ell| < \epsilon$ is unbounded: Given any $N$, there is such an $n$ above $N$. Now a set of natural numbers is bounded if and only if it is finite.