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The definition (from my book) we take is:

$x$ is a limit point of $(x_n)$ if $\exists$ a subsequence $(x_{n_k})$ of $(x_n)$ so that $\lim_{k \to \infty} x_{n_k} = x $

Now, this how I understand equivalences to this definition and its negation:

${\bf I.}$ $x$ is limit point of $(x_n)$ if $\forall \epsilon >0$ $\exists\ N > 0$ so that $k > N$ implies $|x_{n_k} - x | < \epsilon $.

${\bf II}.$ If we put put $X = \{ x_n : n \in \mathbb{N} \}$ if for any $\epsilon > 0$, we have infinitely many of the $x_n's$ lying in $B_{\epsilon}(x)$

Now, these are equivalent to the definition of a limit point.

The negation would be: $x$ is ${\bf not}$ a limit point if $\exists\ \epsilon >0$ such that for all $N > 0$ one can find $k > N$ so that $|x_{n_k} - x| \geq \epsilon $

Is my understanding correct?

Sahiba Arora
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James
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  • Yes. I ve used this many times – Grada Gukovic May 19 '20 at 06:03
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    ${\bf I.}$ Does not make sense. What is $x_{n_k}$ in this statement? – Kavi Rama Murthy May 19 '20 at 06:05
  • IT is a subsequence – James May 19 '20 at 06:07
  • That is not a proper definition of limit point. – Kavi Rama Murthy May 19 '20 at 06:24
  • Kavi: Well, that is why I am asking here. I wrote my understanding, and I Wanted to see whether I am understanding it correctly. What would then be a proper definition of limit point? – James May 19 '20 at 06:26
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    @James: Sahiba Arora’s answer has a nice, careful version of what you were trying for. The main problem with yours is that you never actually mentioned the subsequence: its element $x_{n_k}}$ just suddenly appears out of nowhere. Insert ‘$(x_n)$ has a subsequence $(x_{n_k})$ such that ’ after ‘if’, and you’ve have a less careful but still adequate version of Sahiba Arora’s. – Brian M. Scott May 19 '20 at 16:15

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${\bf II}.$ seems right but ${\bf I}.$ needs to be modified as follows:

${\bf I'}. x $ is a limit point of $(x_n)$ if there exists a sequence $(n_k)_k$ in $\mathbb{N}$ such that $n_k \to \infty$ and for each $\epsilon>0,$ there exists $N \in \mathbb{N}$ such that $k>N$ implies $|x_{n_k}-x|<\epsilon.$

The negation would also need to tweaked as follows:

$x$ is not a limit point of $(x_n$) if for every sequence $(n_k)_k$ in $\mathbb{N}$ with $n_k \to \infty,$ there exists $\epsilon >0$ such that for every $N \in \mathbb{N},$ there exists $k>N$ with $|x_{n_k}-x|\geq \epsilon.$

Sahiba Arora
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