The following are examples of subsets of real numbers *with* accumulation points:

- The real sequence $\left(\frac 1n\right)_{n > 0}$ has the accumulation point $0,$ since for any $\epsilon > 0$ there is an index $n\in\mathbb N$ such that $|1/n – 0| < \epsilon.$
- By definition of convergence, the limit of every convergent sequence is at the same time its accumulation point.
- If $[a,b]$ is a real interval, then any real sequences $(x_n)_{n\in\mathbb N}$ with $x_n\in[a,b]$ is bounded, and contains according to the theorem of Bolzano-Weierstrass a convergent subsequence. Thus, any such sequence $(x_n)_{n\in\mathbb N}$ has an accumulation point.
- All points of the real interval $[a,b]$ are its accumulation points.
- All rational points in the real interval $[a,b]\cap\mathbb Q$ are its accumulation points.

The following are examples of subsets of real numbers *without* accumulation points:

- The set of natural numbers $\{0,1,2,\ldots\}$ has no accumulation points.
- The set of integers $\{\ldots,-2,-1,0,1,2,\ldots\}$ has no accumulation points.

| | | | created: 2017-08-19 21:55:00 | modified: 2017-08-19 22:35:05 | by: *bookofproofs* | references: [6823]

[6823] **Kane, Jonathan**: “Writing Proofs in Analysis”, Springer, 2016