# Prove that the limit of a convergent sequence is an accumulation point.

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 $$S=F\cup(S\setminus F)$$.
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 $$R$$.
• 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. – Abhi May 7 '20 at 20:45