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!
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$\begingroup$ Bounded subset of $\mathbb{R}$. $\endgroup$– NosratiCommented Mar 5, 2017 at 18:12
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2$\begingroup$ the naturals are a counterexample $\endgroup$– The CountCommented Mar 5, 2017 at 18:20
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2 Answers
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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.
answered Mar 5, 2017 at 18:23
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1$\begingroup$ in my book is written that each infinite set of R has at least one accumulation point so that's why i asked can you explain me how is it possible whereas Z is a subset of R it is infinite but it still has no accumulation point $\endgroup$ Commented Mar 5, 2017 at 19:30
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1$\begingroup$ As you've stated it, the book is incorrect (though I can't speak for whether you've quoted it accurately). $\endgroup$ Commented Mar 5, 2017 at 20:55
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$\begingroup$ Is the example I gave for Z as infinite set a good example that shows that there's a mistake in that sentence in my book? $\endgroup$ Commented Mar 5, 2017 at 21:02
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$\begingroup$ Without further context, I can't guarantee e.g. that you haven't just misread it. The "theorem" you quoted is incorrect. If you quoted it accurately and the book contains no other context, then the book is incorrect. $\endgroup$ Commented Mar 5, 2017 at 21:16
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2$\begingroup$ The book is correct. $\mathbb{Z}$ has both $\pm \infty$ as its accumulation points in $\overline{\mathbb{R}}$. (Indeed, we can find an infinite sequence drawn from $\mathbb{Z}$ whose limit is $\infty$: one such sequence is $a_n = n$.) $\endgroup$ Commented Mar 5, 2017 at 22:28
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Every infinite bounded closed subset $S \subset \mathbb R$ has an accumulation point.
For example, if $S = [0,1]$, then all of the points in $S$ are accumulation point.
If $S= \{ 0 \} \cup \{1, \frac 1 2, \frac 1 3, \frac 1 4, \frac 1 5, \dots \}$, then $S$ has one unique accumulation point at $0$.
Your example $S = \mathbb Z$ is not bounded, so the theorem does not apply.
answered Mar 5, 2017 at 18:51
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$\begingroup$ yeah I know that Z is not bounded that's why i asked because in my book it's written that each infinite set in R has at least one accumulation point and I found that Z is an infinite set of R but still has no accumulation point so I'm asking is it written wrong in my book? $\endgroup$ Commented Mar 5, 2017 at 19:44
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$\begingroup$ can you please respond to my question??? $\endgroup$ Commented Mar 5, 2017 at 19:57
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2$\begingroup$ Hi Linda, I think it is witten wrong in the book. Your example of the integers is a perfect counter-example. I hope this helps. $\endgroup$ Commented Mar 5, 2017 at 21:24
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$\begingroup$ Okay thank you for responding to my question I really appreciate it! $\endgroup$ Commented Mar 5, 2017 at 21:28
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$\begingroup$ No problem. Welcome to math.stackexchange, and thank you for your nice question! $\endgroup$ Commented Mar 5, 2017 at 21:29