Re: infinity
- From: Kirby Cook <kwmcook@xxxxxxxxxxx>
- Date: Tue, 09 Aug 2005 20:49:05 GMT
William Hughes wrote:
<snip>Tony Orlow (aeo6) wrote:
William Hughes said:
This is true, but you only add 10 balls at a finite step. No, you never get the empty set at any finite step. No one disputes this.
You never get an empty set at ANY step.
-You never get the empty set at any finite step -we only have finite steps -you never get an empty set at ANY step
looks like we agree. What I also have is the concept of the state after all the finite steps. Call this state, state E Which balls are in the vase at state E. We do not know, this has to be defined.
I balk at "*all* the finite steps". Easy to say, hard to define.
Actually, I - O, right? And that is not at E (which has yet to be defined, IMO) but at any given specific step.
Definition 1:
Let I be the union of the sets of balls added to the vase at any finite step.
Let O be the union of the sets of balls removed from the vase at any finite step.
Then 0 is a subset of I and we define the set of balls that is in the vase at state E is the set difference I\O.
How many balls are in the vase at state E?
Definition 2:
Define the number of balls in the vase at state E to be the number of balls in the set I\0.
Now you may not like these definitions, but they seem natural to me. Furthermore, any definitions that are inconsistent with these seem unnatural to me.
Using the above definitions
1. The vase is empty at state E
2. The number of balls in the vase at state E does not depend on the number of balls in the vase at any finite step (in particular the number of balls in the vase at state E is not any sort of infinite sum).
You may not like the conclusions, but if so you need to disagree with the definitions.
At EACH step, when you remove 1 ball, you have just added 10 balls. Unless, of course, you are saying that at infinite steps you are following a different procedure, but that was never stated. So, you can NEVER, at any finite OR infinite step, have an empty set.
Nothing was said about infinite steps because they do not exist.
here they are arguing that there is a difference between whether the one ball removed is labeled one way or another. If we add 10x through 10x+9 (for x=0 to N) and then remove 10x, then 10x+1 through 10x+9 are NEVER removed. However, if you say you are adding 10x through 10x+9, and taking away x+1, then you claim that you remove all elements. There is a serious inconsistency here between claiming that order doesn't matter for sets in general, and then claiming that here it does. What difference does the label on the removed ball really make?
Without "labels" or other identifying features, you can't decide whether a given ball, once put in, is taken out again at some time.
It doesn't matter about any particular ball. Whenever you remove a ball, you have just added 10, so the vase can't be empty. QED
But you only remove a ball af a finite step. So you have just concluded that the base is not empty at any finite step.
Then you never remove balls at an infinite step? How can it ever become empty then? Do you also stop adding balls at infinite steps? It sounds like infinite steps don't even exist, but that seems like par for the course with the current misunderstanding of infinity.
Check the definitions. I have no need of the hypothesis that infinite steps exist. (In other contexts it is convenient to call state E, step omega, but I don't have to do that here.)
Since it is feasible to remove just a subset (we are talking about infinite sets here), there is no way except checking the balls. I could equally well forget ball 0 and start the removal action with ball 1. In that case, at the end ball 0 will be left. I could remove just the odd-numbered balls, in which case the even-numbered balls will be left. And so on.
So, the same operation results in entirely different numbers of remaining balls, depending on the labels on the balls? This is sheer nonsense.
The solution to the problem lies in forgetting about your bijections, and noting that at each of an infinite number of steps you are adding a net 9 balls to the vase,
Oh, but counting to 9 is all about bijections.
It's about counting. What purpose is this statement supposed to serve?
and the overall number never decreases from one step to the next. You see, it's not that I am missing anything, but simply noting that your system produces contradictory results, none of which make very good sense.
Once you realize that an infinite set can be mapped 1:1 to a proper subset of itself, it becomes clear that at the "end of the process", we can no longer rely on the pigeon hole principle: the number of things put in and taken out alone does not suffice for determining anything (well, it is sufficient for determining that the remaining number of balls must be countable, but that is not really impressive).
When every removal is accompanied by a larger addition, the vase will never be empty, but continually more full. Does the infinite series (10,-1,10,-1,10....) converge to zero? Do the terms have a limit of zero? No, sorry.
Does the problem have anything to do with infinite series. No sorry.
Sorry, but the problem has EXACTLY to do with a sum of an infinite series of terms, which is what infinite series IS. Otherwise, what do you think infinite series are all about? Do you, or do you not, add 10, then remove 1, as described above, an infinite number of times? Are you not summing an infinite number of terms, and if so, don't the rules regarding infinite series apply?
No check the definitions above. I am looking at the number of elements in a set difference. I am not using any sort of infinite series.
<snip>
-William Hughes
Kirby .
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