
Re: Another AC anomaly?
Posted:
Dec 12, 2009 11:22 PM


"K_h" <KHolmes@SX729.com> writes:
> "Jesse F. Hughes" <jesse@phiwumbda.org> wrote in message > news:87vdgcksy2.fsf@phiwumbda.org... >> "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 = {}. > > So, you're claiming that he is not using { and } just to > bracket the argument (i.e. the X_n to be limited) but {X_n} > refers to a set containing the one set X_n.
Er, yes. Though, something seems to be wrong with your notation. At issue is the set X_n = {n}, not the set {X_n}.
In summary:
lim X_n = lim {n} = {} = 0.
lim X_n = lim {n} = lim 1 = 1.
> Then it seems like the meaning has changed because in previous posts > he writes that limS_n=/=limS_n follows from a wikipedia > definition applied to sequences of natural numbers n  not to the > nonnaturals {n}. For example: > > > > I have explicitly defined the limit of a sequence of > sets. With that > > > definition (and the common definition of limits of > sequences of natural > > > numbers) I found that the cardinality of the limit is > not necessarily > > > equal to the limit of the cardinalities.
And that's absolutely correct, as we see above.
> Okay, if {X_n} refers to a set containing the single set X_n > then lim(n>oo){n} is not a limit of the natural numbers > since the naturals are not the sets {n} but the sets n.
Er, yes. Of course.
> In this case my proof shows that lim(n>oo)n=N. Applying the > wikipedia definitions to n is sensible but applying them to {n} > makes a mockery of the notion of a limit.
You have some very odd notions yourself. It's a simple application of a perfectly sensible definition of limit.
> The basic idea behind a limit is that things in one state tend to > some final state and a good definition and application of a limit > should embody that. In looking at the sequence {n}, with 0={} and > 1={0}, saying that it tends to 0={} is a betrayal of the core idea > behind a limit: > > 1, {{0}}, {{0,1}}, {{0,1,2}}, {{0,1,2,3}}, ... > 0 > > The basic idea of what a limit is suggests that an > appropriate definition for lim(n>oo){n} should yield > lim(n>oo){n}={N}: > > {}, {{0}}, {{0,1}}, {{0,1,2}}, {{0,1,2,3}}, ... > > {{0,1,2,3,4,...}} > > In other words, applying the wikipedia definitions to {n} is > an abuse of those definitions. The definition that is used > for a limit should make sense for the kind of object it is > applied to.
You're welcome to your own cockamamie opinions about whether a particular definition is sensible or not, but they're utterly irrelevant to the issue at hand. The fact is that with this *perfectly standard* definition of limits, we see that
lim X_n != lim X_n.
That's all there was at issue.
 "Sorry, wakeup to the real world. You're on your own dependent on me as your guide. Luckily for you, I'm selfcorrecting to a large extent, so if the proof were wrong, I'd tell you. It's not wrong."  James Harris confirms that his proof is correct.

