Based in the definition of the accumulation point is says that: A point $x$ is said to be accumulation point in $S$ if every open set containing $x$ contains at least one other point from $S$. So if $\Bbb{Z}$ (set of all integers) is in $\Bbb{R}$ (set of all real numbers) where $\Bbb{Z}$ is infinite set but I can't find any accumulation point because we can find open set for every integer that contains no other integers therefore all integers are isolated points. I don't understand the mistake in my example. Can someone explain it to me. I would appreciate it!
[Math] Is it true that each infinite set in R has at least one accumulation point
real-analysis
Best Answer
I think your question is: "$\mathbb{Z} \subset \mathbb{R}$ is an infinite set but it has no accumulation points. What's wrong with this statement?"
The answer is that there is nothing wrong with this statement. The integers are a counterexample to the claim that all infinite subsets of $\mathbb{R}$ have an accumulation point.
Apparently your book claims that $\pm \infty$ are allowed as accumulation points. In that case it is correct. (From your question's original wording, I assumed you were working in the reals rather than in the reals with infinity.) Indeed, take any infinite subset of $\mathbb{R}$. Either it is bounded, in which case it is contained in a closed bounded interval and so has an accumulation point by Bolzano-Weierstrass, or else it is unbounded, in which case for every $n$ we can pick an element of the set which is of magnitude bigger than $n$. The resulting sequence demonstrates that at least one of $\pm \infty$ is an accumulation point.