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  1. Suppose $\exists$ function $f: \mathbb{N} \to \mathbb{N}$ s.t. as $n \to \infty$, $f(n) \to \infty$.

Prove that $\forall$ events (or sets) $A_1, A_2, ..., \limsup A_{f(n)} \subseteq \limsup A_n.$

  1. Suppose $\exists$ events (or sets) $A_1, A_2, ...$ s.t. $\limsup A_{f(n)} \subseteq \limsup A_n$ for some function $f: \mathbb{N} \to \mathbb{N}$.

Prove that $f(n) \to \infty$ as $n \to \infty$.

Context: This arises from here.

Here is what I have tried:

1

Suppose $\exists K \in \mathbb{N}$ s.t. $f(n) \to K$.

Does it follow that $f(n) = K$? I can't seem to think of a counterexample.

Anyway, if so $\omega \in A_K \notin \limsup A_n$

If not, I think such will be the case anyway?

2

We want to show that

given $\omega \in \Omega$, if $\omega \in \limsup A_{f(n)}$ and $f(n) \to \infty$, then $\omega \in \limsup A_n$.

This means by definitions that

if $\forall M > 0, \exists N > 0$ s.t. $f(n) > M$ if $n > N$, and $\forall m \in \mathbb{N}, \exists n_1 \geq m$ s.t. $\omega \in A_{n_1}$, then $\forall m \in \mathbb{N}, \exists n_2 \geq m$ s.t. $\omega \in A_{f(n_2)}$.

I think such $n_2$ exists and that it is the natural number s.t. $f(n_2) > M$ if $n_1 > N$ or something like that?

Community
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BCLC
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1 Answers1

2

If $A_n=A$ for each $n\in\mathbb N$ then evidently $\limsup A_n=A$.

For any function $f:\mathbb N\rightarrow\mathbb N$ we have $A_{f(n)}=A$ for each $n\in\mathbb N$ so that also $\limsup A_{f(n)}=A$.

Then $$\limsup A_{f(n)}=A\subseteq A=\limsup A_n$$ but no conditions like $f(n)\rightarrow\infty$ on function $f$ are necessary.

drhab
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  • 1 What do you mean by this? It seems like you are proving such only on the assumption that all of the sets in the sequence are equal to each other? 2 Perhaps such conditions are not sufficient or necessary. I am just wondering how to use and infer $f(n) \to \infty$ ^-^ – BCLC May 22 '15 at 13:10
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    @BCLC This answers the question as it is written (if you have interpreted it differently, you need to rewrite it). In your question, the ${A_n}$ and $f$ are given, so the implied quantifiers are "for all sequences ${A_n}$ and all functions $f$ the below equivalence holds". The answer shows that there exists a choice of ${A_n}$ which makes the LHS true regardless of $f$, meaning it cannot be equivalent to the RHS (since one can easily construct an $f$ for which $f(n) \not\to \infty$). – Erick Wong May 24 '15 at 16:55
  • @ErickWong THANKS! Hehe. So how about now? – BCLC May 24 '15 at 17:27
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    @BCLC After the edit, the implied quantifiers are basically the same. The same choice of ${A_n}$ makes the LHS true regardless of $f$, but the RHS is false for some $f$. – Erick Wong May 24 '15 at 18:03
  • @ErickWong I removed the iff stuff. How about now? – BCLC May 24 '15 at 18:32
  • @BCLC Looking more reasonable, but note that the same counterexample yet again works for your question 2. – Erick Wong May 25 '15 at 03:49
  • @ErickWong Well, one direction is sometimes trivial to prove? http://math.stackexchange.com/questions/924865/on-for-all-in-if-and-only-if-statements-in-probability-theory-and-stochastic-c – BCLC May 29 '15 at 10:30
  • So I guess @drhab did only one direction? – BCLC May 29 '15 at 10:32
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    In this answer I proved that $f(n)\rightarrow\infty$ is not necessary for $\limsup A_{f(n)}\subseteq\limsup A_n$. Here I proved that it is sufficient. In the original question (before editing) you were only asking for one side. Btw, do not change questions essentially by editing. This to avoid making answers irrelevant. – drhab May 29 '15 at 10:39