Date: Dec 12, 2009 12:23 PM
Author: Jesse F. Hughes
Subject: Re: Another AC anomaly?

"K_h" <KHolmes@SX729.com> writes:

> The only way lim(n ->oo){n}={} is if limsup and liminf both
> equal {}. If limsup and liminf are different then the limit
> does not exist and cannot equal an existing set like {}.
> Since the empty set exists, and since you are claiming that
> lim(n ->oo){n}={}, you need to show that limsup and liminf
> are both {} by the definition you are using.


Yes, he needs to show that, but it is utterly trivial and obvious from
the definition of limsup and liminf given here. I'm not sure why you
think it's not obvious, but here's the proof.

Now, let X_n = {n}. Thus, n is in X_k <-> n = k.

n in lim sup X_k iff n is in infinitely many X_k, but we see from the
above that n is in only one X_k. Thus, lim sup X_k = {}.

n in lim inf X_k iff there are only finitely many X_k such that n not
in X_k, but again, we see that this is false for every n. Hence
lim inf X_k = {}.


--
Jesse F. Hughes
"It is a clear sign that something is very, very, very wrong, as human
beings are, well human. Maybe some people think that mathematicians
are not, but I disagree. They are human beings." -- James S. Harris