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What kind of $f(n): \mathbb{N} \to \mathbb{N}$'s make the ff statement true? What kind don't?

$\limsup A_{f(n)} \subseteq \limsup A_n$ where $n \in \mathbb{N}$ (*)

Well obviously the answers to each are:

$(f(n) \ | \ \limsup A_{f(n)} \subseteq \limsup A_n)$

and

$(f(n) \ | \ \limsup A_{f(n)} \subsetneq \limsup A_n)$

respectively, but I am wondering about some possible subsets of each.

For instance

$3n^2 + 5n \in (f(n) \ | \ \limsup A_{f(n)} \subseteq \limsup A_n)$?

or in general

$(f(n) = \sum_{i=0}^{m} a_in^i\ | a_in^i \in \mathbb{N}, m \in \mathbb{N}) \subseteq (f(n) \ | \ \limsup A_{f(n)} \subseteq \limsup A_n)$? If this is right then I guess all polynomial functions whose range $\in \mathbb{N}$ make (*) true.

What if $m = \infty$?

What about continuous functions whose range $\in \mathbb{N}$?

What about some counterexamples to (*)? I think a simple piecewise or discontinuous function will do, but I can't think of any.

Martin Sleziak
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BCLC
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    is $f$ a function $\mathbb N\rightarrow\mathbb N$ here? – drhab Sep 12 '14 at 09:26
  • @drhab Yes. Sorry for the confusion. Will edit. Thanks. – BCLC Sep 12 '14 at 09:27
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    Necessary and sufficient condition: $f(n)\to\infty$ when $n\to\infty$. – Did Sep 24 '14 at 06:39
  • "This question has not received enough attention." It happens it has, see my previous comment. – Did Oct 07 '14 at 18:18
  • @Did I totally did not see that answer. Thanks. Could you explain further? – BCLC Oct 07 '14 at 18:19
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    No idea about one sense of the implication? Or the other? Now that you realized the existence of this comment, you might want to spend one minute or two pondering why it holds. – Did Oct 07 '14 at 18:21
  • @Did Fiiiine. Hmmm...how do you know it is necessary? Got sufficiency. :) – BCLC Oct 07 '14 at 18:24
  • Necessity is even more trivial. THINK. – Did Oct 07 '14 at 18:25
  • @Did Oh I think I mixed them up. How do you know it is sufficient? :P Wait. Ugh. "How do you know (*) implies your comment?" --> That is asking sufficiency? – BCLC Oct 07 '14 at 18:27
  • I was just wondering, what are the (An)? Any particular property the sequence verifies? – mvggz Oct 10 '14 at 17:39
  • @mvggz nah. Just a bunch of sets – BCLC Oct 10 '14 at 18:50

1 Answers1

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Not a full answer to your question, but maybe this can help.

For any $x$ define $A_{x}:=\left\{ n\in\mathbb{N}\mid x\in A_{n}\right\} $ and $B_{x}:=\left\{ n\in\mathbb{N}\mid x\in A_{f\left(n\right)}\right\} $.

Then $x\in\limsup A_{n}$ if and only if $A_{x}$ is infinite, and $x\in\limsup A_{f\left(n\right)}$ if and only if $B_{x}$ is infinite.

Note that $n\in B_{x}\iff x\in A_{f\left(n\right)}\iff f\left(n\right)\in A_{x}$ showing that $B_{x}=f^{-1}\left(A_{x}\right)$.

Now we can say that $\limsup A_{f\left(n\right)}\subseteq\limsup A_{n}$ is true iff the implication: $$A_{x}\text{ is finite}\Rightarrow f^{-1}\left(A_{x}\right)\text{ is finite}$$ is true for each $x$.

It is immedeately clear that $f$ will satisfy this condition if it is injective and this can be weakened.

drhab
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  • Why did you withdraw your acceptance? – drhab May 06 '16 at 16:04
  • drhab, well it wasn't a full answer... – BCLC May 06 '16 at 16:08
  • Well, in the future then only accept full answers. Of course it is up to you what to accept or not, but be careful with it, so that the number of withdrawals will tend to zero. I am a bit displeased about this. Especially because it was more than a year ago that the answer was accepted. – drhab May 06 '16 at 16:17