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From this question: Constructing a set with exactly three limit points.

To answer this question

Construct a bounded set of real numbers with exactly three limit points.

But why $(1,2)\bigcup (2,3)$ can't be answer?

From my understand to limit point, $(2,3)$ only have two limit point $2$ and $3$. Any $x$ not be $2$ or $3$, can't be $(2,3)$'s limit point. Am I right?

Community
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jiamo
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1 Answers1

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No.

If you understand why $(2,3)$ is an open set then,

Lemma: Let $X$ be an open set in $\mathbb{R}^k$ , if $p\in X$ then $p$ is a limit point of $X$. (Hint: every neighborhood of $p$ contains an element other than $p$)

Thus every point in $(2,3)$ is a limit point. Also we have two additional limit points, namely $2$ and $3$.

hrkrshnn
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  • I got it, relook the define of limit point. Since I think the limit point should at boundery of a set , that mean it is limit to the boundery but can't just array. – jiamo Sep 07 '14 at 13:54
  • I may mean I think a limit point like a limit value of a f(x) when x -> $\infty$. – jiamo Sep 07 '14 at 14:19