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Prove that a number a is a cluster point of a set A iff for any $\delta > 0$, the interval $(a-\delta, a+\delta)$ contains infinitely many points of A.

My work:

I found a proof for one direction in case of $\mathbb{R^2}$ in schaum series "Theory and problems of general topology" in page 56 problem no.7 but it seems to me that in this way the solution of the above question would be very long, could anyone suggest a smarter solution for me please?

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  • $\begingroup$ Can you spell out the definition of cluster point you are using? $\endgroup$ – Kavi Rama Murthy Oct 9 '18 at 6:11
  • $\begingroup$ yes sure@KaviRamaMurthy $\endgroup$ – Idonotknow Oct 9 '18 at 6:12
  • $\begingroup$ we say that a numbe ra is a cluster point of a set A, if every interval$(a-\delta, a+\delta)$ contains at least 1 point of A, not counting a. @KaviRamaMurthy $\endgroup$ – Idonotknow Oct 9 '18 at 6:17
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Suppose $(a-\delta, a+\delta)$ contains only finite number of points of $A$ other than $a$, say $\{a_1,a_2,...,a_N\}$. Then there exists $\delta'$ such that $(a-\delta', a+\delta')$ contains no point of $A$ other than $a$, leading to a contradiction. Take, for example, $0<\delta' < \min \{|a_i-a|: 1\leq i \leq N\}$. The converse implication is immediate from definition.

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  • $\begingroup$ Is this a proof for the 2 directions? $\endgroup$ – Idonotknow Oct 9 '18 at 6:39
  • $\begingroup$ @Idonotknow I have added another line to my answer. $\endgroup$ – Kavi Rama Murthy Oct 9 '18 at 6:42
  • $\begingroup$ So this small proof is the whole answer? ... wow! $\endgroup$ – Idonotknow Oct 9 '18 at 6:45
  • $\begingroup$ does contains at least one point of $A$ implies contains infinitely many points of $A$? $\endgroup$ – Idonotknow Oct 9 '18 at 8:38
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    $\begingroup$ No. the other way. Containing infinitely many points implies containing at least one point. $\endgroup$ – Kavi Rama Murthy Oct 9 '18 at 8:40

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