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Prove: a point a is a cluster point of a set $A \subset \mathbb R$ iff there exists a sequence ${a ^{(k)}} \subset A\setminus\{a\}$ converging to $a$.

My thoughts:

I know that the definition of a cluster point $a$ of a set $A \subset \mathbb R$ is, for every $\delta > 0$, the n-ball $B_{\delta}(a)$ contains at least one point of A, not counting $a$. but I do not know how to use this definition to prove what required.

Could anyone show me how to prove this please?

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    $\begingroup$ Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for. $\endgroup$ – 5xum Feb 19 at 9:14
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    $\begingroup$ @5xum: I'm not sure why the change of title. $\endgroup$ – Asaf Karagila Feb 19 at 9:18
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    $\begingroup$ It would be good if you include the definition of cluster point you're working with in the question. $\endgroup$ – coffeemath Feb 19 at 9:18
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    $\begingroup$ @5xum: The goal of a title is to tell you what is the question, as best as possible (with the occasional clickbait in some rare occasions). Yes, the question shouldn't be just in the title. But there's no problem with a title saying exactly what is the characterization sought for. "How to solve this problem" is very nondescript and very annoying compared to "Proving that X is equivalent to Y". $\endgroup$ – Asaf Karagila Feb 19 at 9:30
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    $\begingroup$ @5xum: I never said "always include the question in the title". $\endgroup$ – Asaf Karagila Feb 19 at 9:31
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Hint:

Take $\epsilon = \frac{1}{n}$ in the definition of a cluster point to find a sequence. For the other direction, the fact that a sequence in $A \setminus \{a\}$ exists already tells you something about the intersection of open balls with $A$.

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  • $\begingroup$ Could you please include some details about the other direction? $\endgroup$ – Secretly Feb 19 at 9:32
  • $\begingroup$ @hopefully For the other direction, you want to show that every ball with an arbitrary radius $\epsilon>0$ intersects $A$ at a point other than $a$. Right? This means that you should show that $B_r(a) \cap (A\setminus \{a\}) \neq \emptyset$. Now, what does the definition of $\lim_{n\to\infty} a_n = a$ tell you about this? Think about the terms of the sequence $a^{(k)}$. By the way, the notation $a^{(k)}$ to denote a sequence is kind of nonstandard. To avoid confusion, use $\{a_k\}$ to denote a sequence indexed by the dummy variable $k$. $\endgroup$ – stressed out Feb 19 at 9:36
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    $\begingroup$ Why am I sure that I will find $x_n$? Because by definition, given any $\epsilon>0$, the set $B_r(a)\cap (A\setminus \{a\})$ is not empty. For each $n\in\mathbb{N}$, I'm taking $\epsilon = \frac{1}{n}$. Then the definition implies that some point (let's call it $x_n$) must exist in $B_{1/n}(a)\cap (A\setminus \{a\})$. $x_n$ is just a notation for the point that exists in $B_{1/n}(a)\cap (A\setminus \{a\})$ by the definition of $a$ being a cluster point. $\endgroup$ – stressed out Feb 19 at 9:59
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    $\begingroup$ @hopefully Yup. That's correct. $\endgroup$ – stressed out Feb 19 at 10:34
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    $\begingroup$ Thank you so much for your great help. $\endgroup$ – Secretly Feb 19 at 10:37

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