Originally posted by iamatigerThe numbering thing only matters because it ensures every ball is eventually removed. Taking only odd balls does not, so an infinite number of balls remain.
The whole numbering thing is immaterial.
step 1, bag has 1, 2
step 2, bag has 2,3,4
step 3, bag has 3,4,5,6
step 4, bag has 4,5,6,7,8
Even though your friend takes your numbers out, he takes them out too slowly to catch up. Lets do a simple transform that does not change the number of the balls:
The converted ball number on step N is ball ...[text shortened]... hows your friend is getting nowhere. He may as well stop removing balls, it makes no difference.
Your sequence highlights what mtthw already said. At step n (defined as you did), we have balls n, n+1,...,2n. But the lower bound is converging to infinity, yet we know that IF there are any balls left there must be a lowest numbered ball (you're welcome to disprove my argument from before if you think it's wrong), so it's impossible that there are infinite balls in the bag at t=0.
It's still possible that the problem is not well defined, as mtthw said (I don't agree), but it's certainly not possible that the problem is well defined and there are infinite balls in the bag.
Originally posted by PBE6Infinite"If you've done six impossible things this morning, why not round if off with breakfast at Millyways?" - Douglas Adams
Suppose you and your friend come into possession of a bag of infinite capacity, along with an infinite supply of numbered balls and a stopwatch with an infinite number of gradations. After watching the universe explode ...[text shortened]...
[b]Question: How many balls will be left in the bag when the stopwatch reads 0 seconds?[/b]
after 0 secs 2 balls
after 30 secs 3 balls
after 45 secs 4 balls
after 52.5 balls 5 balls
so we have an infinite arithmetic sequence {2,3,4,5, ....}
and there are guys on here saying that the sequence tends to zero?!?!
Originally posted by wolfgang59Yes, there are.
Infinite
after 0 secs 2 balls
after 30 secs 3 balls
after 45 secs 4 balls
after 52.5 balls 5 balls
so we have an infinite arithmetic sequence {2,3,4,5, ....}
and there are guys on here saying that the sequence tends to zero?!?!
Maybe you should read the thread as I've proven that having infinite balls leads to a contradiction. You're welcome to disprove the argument.
Originally posted by PalynkaWould you agree that one way of defining a limit is that one can get as close to the limit as one wants?
Yes, there are.
Maybe you should read the thread as I've proven that having infinite balls leads to a contradiction. You're welcome to disprove the argument.
e.g.#1
1/2+1/4+1/8+ ...
we know limit is 1 because ANY number less than 1 can be achieved in sufficient steps.
(0.9 or 0.9999999 or 0.9999999999999999999999999)
1+2+3+ ...
we know limit is inf because ANY number can be achieved in sufficient steps.
(1,000,000 or 5,000,000 or 9,999,999,999,999,999,999,999)
Now with our current problem how close can we get to zero?
How close can we get to infinity?
Its 30 years since my maths degree so I am rusty. 😞
But I think I can prove the answer is NOT zero.
by definition if the limit of s1, s2, ... is L then for every real number r > 0, there exists a natural number x such that for all n > x, |sn − L| < r.
(This means that eventually all elements of the sequence get as close as we want to the limit)
now if you believe limit is 0
and I give you r=1/2 (not that close!!)
show me x
Originally posted by wolfgang59For every numbered ball N I can find a step S when the ball is removed.
Its 30 years since my maths degree so I am rusty. 😞
But I think I can prove the answer is NOT zero.
by definition if the limit of s1, s2, ... is L then for every real number r > 0, there exists a natural number x such that for all n > x, |sn − L| < r.
(This means that eventually all elements of the sequence get as close as we want to the limit)
now if you believe limit is 0
and I give you r=1/2 (not that close!!)
show me x
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I think an infinite countable set can have finite lower or upper bounds, or neither but not both, and both its bounds can be infinite with the same sign.
The set of balls in this problem has no upper bound, and no finite lower bound but despite its infinite bound with the same sign, and its countability, it is still provably infinite.
For instance it can be proved that the set has an infinite lower bound and its members are equal in number to the integers between 0 and its lower bound.
I suppose the response to anyone claiming there are an infinite number of balls left is "name one".
For me, the root of the problem is this. We know what it means for a sequence of numbers to have a limit - we can define it precisely. But what is the definition of the limit of a sequence of sets?
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Originally posted by mtthwPositive infinity. That's the only name you'll get seeing as both the lower and the upper bound are that. Funny that: An infinite set can have equal lower and upper bounds, and an infinite number of contents nevertheless!
I suppose the response to anyone claiming there are an infinite number of balls left is "name one".
For me, the root of the problem is this. We know what it means for a sequence of numbers to have a limit - we can define it precisely. But what is the definition of the limit of a sequence of sets?