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After going over the definitions of a neighborhood of a point and convergence of a sequence it seems that there is a corollary between both definition (backed with some intuition)

Can we say that if $x_n\rightarrow x$ so $x$ has a neighborhood s.t $x_n\in N_{x}$ if not what is the connection between both definition

gbox
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    When you write $x_n\in N_x$, what is the quantifier on $n$? By the definition of convergence, we know that every neighborhood $N_x$ of $x$ satisfies $x_n\in N_x$ for all but finitely many $n$. – John Griffin Aug 09 '17 at 06:38
  • @JohnGriffin for example we can take $(-0.1,0.1)$ which is a neighborhood of $0$ but $\frac{1}{n}\not{\in}(-0.1,0.1)$ for some $n$ so neighborhood has less "strictly" from the convergence definition? as we can take a big interval. for example all $\mathbb{R}$ is a neighborhood of $0$ – gbox Aug 09 '17 at 06:50
  • Yes, but $(-0.1,0.1)$ still contains all but finitely many terms of the sequence $\frac{1}{n}$, which is all that matters for convergence. Of course bigger neighborhoods will contain "more" elements, but that's true regardless of convergence. The whole space $\mathbb{R}$ is a neighborhood of $100$, so $100$ has a neighborhood containing all terms $\frac{1}{n}$. At this moment I'm not sure what your question is. – John Griffin Aug 09 '17 at 07:14

1 Answers1

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$x_n \to x$

$ \iff$

for each neighborhood $N_x$ of $x$ there is $n_0=n_0(N_x) \in \mathbb N$ such that $x_n \in N_x$ for all $n>n_0$.

Fred
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