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Suppose we have a function $g: \mathbb{N} \to \mathbb{N}$ s.t. $g(n) \to \infty$ as $n \to \infty$.

Is it true that $g \in \{f \ | \ \limsup A_{f(n)} \subseteq \limsup A_n\}$ for some sequence $A_n$?

Suppose g does not approach $\infty$ as $n \to \infty$ where $K < \infty$.

Is it true that $g \notin \{f \ | \ \limsup A_{f(n)} \subseteq \limsup A_n\}$ for some sequence $A_n$?

The sequence $(A_n)$ doesn't satisfy any particular property, apart from $\limsup A_n$ existing which always exists (I think?).

BCLC
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  • Are you sure you mean $g \to K$? This would mean $g(n) = K$ for all $n > n_0$, making the question rather obvious. – Erick Wong Jun 06 '15 at 16:46
  • @ErickWong I am not sure. It is supposed to be the negation of "g(n)→∞ as n→∞." How do I negate that then? – BCLC Jun 06 '15 at 17:30
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    Write out the definition of $g(n)\to\infty$ and negate it. Or just say $g(n)\not\to\infty$. This doesn't mean $g$ has to converge to anything, nor even does $g$ need to be bounded. – Erick Wong Jun 06 '15 at 19:55
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    You need to give a quantifier for $A_n$ in your question. is your RHS the set of all $f$ that satisfy the $\subseteq$ for every choice of $A_n$ or for some choice of $A_n$? – Erick Wong Jun 06 '15 at 19:57
  • @ErickWong Thanks! – BCLC Jun 07 '15 at 08:37

1 Answers1

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On your first question:

Fix some element $x\notin\limsup A_{n}$, wich means:

$$\text{some }n_{x}\in\mathbb{N}\text{ exists with: }\forall k>n_{x}\left[x\notin A_{k}\right]$$

From $\lim_{n\rightarrow\infty}g\left(n\right)=\infty$ it follows: $$\text{some }m_{x}\in\mathbb{N}\text{ exists with: }\forall k>m_{x}\left[g\left(k\right)>n_{x}\right]$$ and consequently: $$\forall k>m_{x}\left[x\notin A_{g\left(k\right)}\right]$$

The last statement tells us that $x\notin\limsup A_{g\left(n\right)}$.

Proved is now that $\limsup A_{g\left(n\right)}\subseteq\limsup A_{n}$ if $\lim_{n\rightarrow\infty}g\left(n\right)=\infty$ .

On your second question.

No, not in general. See here for a proof that counterexamples can easily be constructed.

If e.g. $A_{n}=A$ for each $n$ then $\limsup A_{g\left(n\right)}=\limsup A_{n}$ is true for any function $g:\mathbb{N}\rightarrow\mathbb{N}$.

Community
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drhab
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  • Thanks drhab. As for the second question. so this is wrong? "Necessary and sufficient condition: f(n)→∞ when n→∞." http://math.stackexchange.com/questions/928535/what-kind-of-fns-make-the-limsup-statement-is-true-what-kind-dont#comment1945190_928535 – BCLC Jun 06 '15 at 17:29
  • Yes, that is wrong. It is sufficient but in general not necessary. – drhab Jun 07 '15 at 07:41
  • Thanks, but I think my question may have been wrong. How about now? – BCLC Jun 07 '15 at 08:36
  • If $g\left(n\right)\rightarrow\infty$ then $\limsup_{n\rightarrow\infty}A_{g\left(n\right)}\subseteq\limsup_{n\rightarrow\infty}A_{g\left(n\right)}$ for every sequence $\left{ A_{n}\right} $. If $g\left(n\right)\rightarrow\infty$ is not true then there are sequences $\left{ A_{n}\right} $ such that $\limsup_{n\rightarrow\infty}A_{g\left(n\right)}\subseteq\limsup_{n\rightarrow\infty}A_{g\left(n\right)}$ but also there can be sequences $\left{ A_{n}\right} $ s.t. $\limsup_{n\rightarrow\infty}A_{g\left(n\right)}\subseteq\limsup_{n\rightarrow\infty}A_{g\left(n\right)}$ is not true. – drhab Jun 07 '15 at 10:30
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    Btw, if a question is posted then it is not okay to change it essentially later. That is not neat towards persons who have answered. You (could) make their answers irrelevant. – drhab Jun 07 '15 at 10:32