Let $A=\{1+ \frac{1}{n}$ : n$\in$ $\mathbb N$}. Is $1$ an adherent point, an accummulation point or both for this set? I think that $1$ is an adherent point because $B(1, r) \cap A \neq \emptyset$ for all $r>0$. But what about an accummulation point? Any tips? I don't need a formal proof, just the logic behind it.
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Note: $x$ is an adherent point iff $x \in \overline{A}$. – Ben Grossmann Nov 22 '16 at 20:16
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Your title has accumulated too many 'm's :-). – copper.hat Nov 22 '16 at 20:21
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@copper.hat these accumulation points are particularly delicious – Ben Grossmann Nov 22 '16 at 20:22
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oh, thanks for pointing that out – TanEma Nov 22 '16 at 20:22
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1@Omnomnomnom: Smartie... – copper.hat Nov 22 '16 at 20:23
2 Answers
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For any $\epsilon>0$, there is some $n$ such that ${1 \over n} < \epsilon$ and so $1+{1 \over n} \in B(1,\epsilon)$. Since $1 \neq 1+{1 \over n}$, we see that $1$ is both an accumulation point and an adherent point.
Every accumulation point is an adherent point.
copper.hat
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By definition of an accumulation point, it suffices to note that no matter how small $r>0$ is, we'll never have $(B(1,r) \cap A) \setminus \{1\} = \emptyset$. Since $1 \notin A$, it suffices to note that $B(1,r) \cap A$ is never empty.
Ben Grossmann
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