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Sequence $<x_n>$ is called eventually in $X$ if $\exists n_0\in \mathbb{N}$ such that $x_n\in X,\forall n\geq n_0$

Sequence $<x_n>$ is called frequently in $X$ if $\forall n_1\in \mathbb{N},\exists n\in \mathbb{N}$ such that $x_n\in X \,\text{and}\, n\geq n_1$

My lecture notes say eventually $\implies$ frequently because $\forall n_1\in \mathbb{N}\ni n_1\geq n_0,\exists n\in \mathbb{N}$ such that $x_n\in X \,\text{and}\, n\geq n_1$.

But this prove only works for $\forall n_1\geq n_0$ not $\forall n_1\in \mathbb{N}$. So how could this proof be enough?

Shouldn't the proof be eventually $\implies$ frequently because $\forall n_1\in \mathbb{N},\exists n\in \mathbb{N}\ni n\geq n_0$ such that $x_n\in X \,\text{and}\, n\geq n_1$.

gbd
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    If you can find a $n$ for a $n_1 \geq n_0$ then the same $n$ satisfies the definition of frequently for any $n’_1 < n_0$. So that case doesn’t create problems. – Zanzag Oct 22 '21 at 19:54
  • Would you be more satisfied if the proof read "$\forall n_1\in \mathbb{N},\exists n\in \mathbb{N}$ such that $x_n\in X ,\text{and}, n\geq \max(n_0,n_1)\geq n_1$."? The difference is really just about what we want to emphasize. – Brian Moehring Oct 22 '21 at 20:42
  • @BrianMoehring, Yes this to me is a better proof. Just for clarity is the proof in the lecture notes correct and enough? – gbd Oct 22 '21 at 20:52
  • @gbd It depends what you mean by that. It's correct in every way that matters, but also you are correct if you're claiming that it's not formally complete. On the other hand, very few mathematics proofs are published in a formally complete form. Not even my alternative form is. If you can see how to fill in all the holes, it's complete enough for you. – Brian Moehring Oct 22 '21 at 20:59
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    For what it's worth, once you've worked though all this at the "machine language" symbolic code level to your satisfaction, probably the way you want to think about this later is that eventually means "all but finitely many" and frequently means "infinitely many", at which point this answer may be of interest. – Dave L. Renfro Oct 23 '21 at 11:05

2 Answers2

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Assume that $\{a_n\}$ is eventually in $X$. Let $n_0$ be given accordingly.

Choose any $n_1 \in \mathbf N$. You must show there exists $n \ge n_1$ with $a_n \in X$.

If $n_1 < n_0$ you can let $n = n_0$. Then $n \ge n_1$ and $a_n \in X$.

If $n_1 \ge n_0$ you can let $n = n_1 + 1$. Then $n \ge n_1$, and since also $n \ge n_0$ you have $a_n \in X$.

Umberto P.
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Let $(x_n)$ be eventually in $X$ and let $n_0$ be an integer as assured by the definition. Now consider an arbitrary $n_1 \in \mathbb N$. Define $n = \max(n_0,n_1)$. Then $n \ge n_1$ by definition and $x_n \in X$ since $n \ge n_0$.

Paul Frost
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