Can we say for any given interval in $\mathbb R$ that every point in this interval is accumulation point?

Question

Accumulation point means briefly that if $x_0$ is acc. point in set $S$ then for any given $\epsilon>0$, $((x_0-\epsilon, x_0+\epsilon ) \cap S )\setminus\{x_0\} \not = \emptyset$

My reasoning is, since every (open) interval is open set for every $x$ in the set we can ball containing $x$ inside the set(interval). This means $\epsilon>0$, $((x_0-\epsilon, x_0+\epsilon ) \cap S )\setminus\{x_0\} \not = \emptyset$. Is it wrong then is there any counterexample ?

Can we generalize it to $\mathbb R^n$

The condition for accumulation point is $(x_0-\epsilon,x_0+\epsilon)\cap S$ contains some point $x\ne x_0$. — Eclipse Sun, Nov 08 '18 at 17:35
$\epsilon>0$, $((x_0-\epsilon, x_0+\epsilon ) \cap S )\setminus{x_0} \not = \emptyset$ editted — Micheal Brain Hurts, Nov 08 '18 at 17:37

score 1 · Answer 1 · answered Nov 08 '18 at 17:40

1

Your claim is true. But note that every interval is not necessarily an open set.

answered Nov 08 '18 at 17:40

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Can we say for any given interval in $\mathbb R$ that every point in this interval is accumulation point?

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