So, here's a question from Modern Calculus and Analytic Geometry by Richard Silverman:

A finite number $c$ is called an accumulation point of a sequence $\{x_n\}$ if every neighbourhood of $c$ contains infinitely many terms of the sequence. Prove that if $\{x_n\}$ is a convergent sequence with limit $c$, then $c$ is an accumulation point.

Proof Attempt:

Let $x_n \to c$ as $n \to \infty$. Then, for any $\epsilon > 0$, there exists an integer $N(\epsilon) > 0$ such that:

$$n > N \implies |x_n - c| < \epsilon$$

Now, let $\epsilon_0 > 0$ be fixed. Suppose that there did exist a neighbourhood of $c$, given by $(c- \epsilon_0,c+\epsilon_0)$ which contained only finitely many terms of the sequence.

For this choice of $\epsilon_0$, there exists an integer $N(\epsilon_0) = N_0 > 0$ such that:

$$n>N_0 \implies |x_n - c| < \epsilon_0$$

$$\implies x_n \in (c-\epsilon_0,c+\epsilon_0)$$

Define the set $P = \mathbb{N} - \{1,2,3,\ldots,N_0\}$. So, every term in the sequence that belongs to the neighbourhood above has index that belongs to $P$. It suffices to prove that $P$ is infinite.

Let $f:\mathbb{N} \to P$ be a map defined as follows:

$$\forall n \in \mathbb{N}: f(n) = N_0 + n$$

Clearly, this is bijective. So, there exists a bijective map between $\mathbb{N}$ and $P$, indicating that $P$ is countably infinite. This shows that there are infinitely many terms of the sequence in the neighbourhood above, which is a clear contradiction.

Hence, we have proven that every neighbourhood of $c$ has infinitely many terms of the sequence.

Does the proof above work? If it doesn't, why? How can it be fixed?


It is correct, but it is a waste of time to wast all that time to prove that $P$ is an infinite set. You can use the fact that if $S$ is an infinite set and $F\subset S$ is a finite one, then $S\setminus F$ is infinite. Otherwise, $S$ would be finite, since $S=F\cup(S\setminus F)$.

  • $\begingroup$ Ahhhhhhh, okay, I'm stupid. That is just way more simple. Thank you so much! $\endgroup$ – Abhi May 7 '20 at 20:44

For every neighborhood $R$ of $c$, choose $\epsilon > 0$ such that $B_\epsilon(c) \subseteq R$.

Then, all but finitely many terms of the sequence $x_n$ are inside $B_\epsilon(c)$ and hence, inside $R$.

  • 1
    $\begingroup$ Ahhh, I assume that $B_\epsilon(c)$ is an $\epsilon$-ball centered on $c$? Haha, I've only vaguely encountered that since i have a limited amount of exposure to metric spaces. $\endgroup$ – Abhi May 7 '20 at 20:45

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