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For a finite complement topology on $R$ to which point or points does the sequence ${\frac{1 }{n}}$ converge?

For a finite complement topology on real numbers the only set including the limit point of sequence is $R$ so the sequence converges to every point of $R$. Am I right?

Kavita
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  • What do you mean by "...the only set including the limit point of sequence..." ?? – DonAntonio May 09 '17 at 09:04
  • Basis for finite complement is either finite set or R. There is no finite set containing {0} so R is only set containing it – Kavita May 09 '17 at 09:14
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    No finite sets that contain ${0}$? How about ${0}$ itself? But you probably meant to say that there are no open finite sets that contain ${0}$. That is correct, but not relevant here. I suspect that you are focused on ${0}$ because $0$ should be its limit point. If that's indeed your motivation then your focus is determined by another topology on $\mathbb R$: the usual one (hence your focus makes no sense). In the cofinite topology every point is a limit point of the sequence. – drhab May 09 '17 at 09:46

3 Answers3

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Perhaps the following is what you tried to convey in your question:

Take any point $\;r\in\Bbb R\;$ and let $\;U_r\;$ be any open neighborhood of it, which means that $\;X:=\Bbb R\setminus U_r\;$ is finite and thus there exists $\;M\in\Bbb N\;$ such that $\;m>M\implies \frac1m\notin X\;$, which means that

$$\forall\,m>M\,,\,\,\frac1m\in U_r\implies r=\lim_{n\to\infty}\frac1n$$

DonAntonio
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Let $X$ be a topological space, $\langle x_n:n\in\Bbb N\rangle$ a sequence of points of $X$, and $x\in X$.

Definition: $\langle x_n:n\in\Bbb N\rangle$ converges to $x$ if and only if for each open nbhd $U$ of $x$ there is an $m_U\in\Bbb N$ such that $x_n\in U$ whenever $n\ge m_U$.

Now let $x\in\mathbb R$ be an arbitrary element and let $U$ be an open nbhd of $x$.

Then $U^c$ is finite so some $m_U\in\mathbb N$ exists such that $\frac1n\notin U^c$ whenever $n\geq m_U$.

That can be refrased as: some $m_U\in\mathbb N$ exists such that $\frac1n\in U$ whenever $n\geq m_U$.

So according to the definition sequence $\langle \frac1n:n\in\Bbb N\rangle$ converges to $x$.

Here $x$ was taken arbitrary so $\langle \frac1n:n\in\Bbb N\rangle$ converges to every $x\in\mathbb R$.

drhab
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Yes, any sequence of distinct
elements converges to every point. A sequence with a
finite number of distinct elements, converges iff it's
eventually constant.

William Elliot
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    The first sentence is ambiguously formulated. The sequence $(a_n)$, where $a_n = \pi$ if $n$ is even, and $a_n = \frac{1}{n}$ if $n$ is odd, has an infinite number of distinct elements, but its only limit is $\pi$. You probably meant that all terms of the sequence should be distinct. – Daniel Fischer May 09 '17 at 10:56