I tend to get conflicting answers as to whether 2^countable gives an answer that is of countable or uncountable cardinality
It technically depends on ordinal versus cardinal exponentiation
Essentially though I am asking as to whether 2 4 8 16,... etc limits to a certain uncountable number----- and if the answer is that it isnt, is this simply because ordinal standard arithmetic does not have a concept of uncountable numbers and only works with a limit concept such that for any arbitrarily large finite number the value is an integer is thus cannot be uncountable (lest there be an integer designating a cardinality of the continuum which would mean that there is an integer bigger than all integers contradiction). So when ordinal arithmetic gives the answer as countable is this just an ambiguity in language or due to a concept that cannot express the continnuum. For instance storrs mccall express a decenary tree concept in his 1994 book (a model of the universe) in which there are ten branch which split in ten branches at time =t + t/2, split again into 1000 at t+ 3t/4, such as it limits to 2t we get uncountably many branches at the end.Does this presume some kind on non standard analysis to get actually get to the concept of reaching the limit, or is his claim just false that there would uncountably many branches-- there is for any real number an binary (and or decimal-decanary) expansion in terms of a countably infinite long sequences of integers between 0-1/0-9, and so it would presume that therefore there are uncountably many such countable sequences indicating that the number of sequences 10-ary sequences of countably infinite length, is uncountable, Is this correct, and if this cant this be expressed roughly as a branching splitting process, where each branch is a digit from 0-9, for countably many such splits.
I also ask this because st petersburg paradox relies on a utility that is of the form 2^x....... for a countably infinite long sequence x, which would appear to indicate one is attempting fit an uncountable utility within such sequences; and conversly and probability values whose values 1/(some uncountable number) which could not be expressed as a relative frequency in a countably infinite long sequence
You are confusing at least three different concepts here. First, the sequence of integers {1, 2, 4, 8, ...} does not have a limit with the usual (Euclidian) metric. It is possible to define a metric, so that the sequence becomes Cauchy, and it is possible to define a new entity, call it "oo" that represents this limit. It is a single entity, hence it is of course countable (but it is not an integer). (If you think of {1, 2, 4, 8, ...} as sequence of real numbers, you get the same result.
Second, the idea of "branches" in the infinite tree is fuzzy. (This may not be your mistake, but it needs to be addressed nonetheless.) Take first the finite tree
o
/ \
o o
/ | | \
o o o o
I hope this is readable; I am not familiar with this interface. The idea is to have a single node at the top, two nodes branching from it, and then two more nodes branching from each of them. How many branches are there? Three or four? The answer depends on what you call a "branch". If you refer to the act of branching, then there is one branch at the top node, a second one at the node to the left below it, and a third branch to the right. The four terminal nodes at the bottom of the picture are properly called "leaves", not branches, to underscore the distinction. You also have the paths that lead from the root node (at the top) to each of the four leaves. There is a natural mapping from leaves to paths and vice versa.
Then, go to the infinite tree. Note that the infinite tree *does not have leaves*. Each *path* is an infinite sequence of left and right branches (it never stops), and there is an uncountable number of paths. However, the set of *branches* --- in the sense I used them above --- is countable! There is one branch at the top level, two at the next, followed by four, eight, etc.
This is an example of a situation where lim f(...) =/= f (lim ...)
well i am talking about the number of paths, not branches. also when you mean branches do you at any point left or right or counting them all (and do you count each of the branches from left to right as a branching or each branch-- so in your example you could define the number of branchings as three if you count the first and the two at the next level (so now its a limit of a sum of 1 +2 +4, not the limit of that series itself), or you could just mean the number of branches at the top level left to right which wouldd be 2 or you mean just 'branching' which is just the number of node splits which is just one at every level (i can see that this is countable)- however under the other two defintions it the total number of branchings would appear to equal almost to the number of leaves (ie either n-1, if there are 8 leaves there are 7 branching counted top to bottom, or othwise just counting the number at the top, 4 branchings which would always be half the number of paths/ leaves, I mean i can see that the number of leaves, at any top node, progress from 1 to 2 (after one branching) to 4 (after two branching etcs)..
Now by paths do you count entire paths (entire histories), or at each level summing them up. I generally mean the former, in which case there is a 1-1 correlations between entire paths and leaves at teh top node at any point in time. And it appears to be the number of entire paths/histories (as i call them) can be seen to limit from 2- 4 to 8 etc. In what sense can there be an uncountable number of paths, when it corresponds to the limit of 2, 4, 8 and yet this sequence does not constitute an uncountable number that is being limited towards- of course in ordinal arithmetic we never reach the limit and as a limiting concept, we consider what the entity is at every point in arbitrarily long sequence, and it always an integer and is countable so we use induction, as it were, to deduce that the limit is also countable but perhaps such an induction is unwarranted, it is unwarranted to call it countable just as much as it is unwarranted to say it even has a limit.
Is it just the ambiguities of the different concept when applied in cardinal arithmetic as actual infinities, versus ordinal arithmetic that only deals with limits. I mean could we use nonstandard analysis to capture the entire infinite sequence 1, 2,4, etc, sum up every one of the numbers, where we know that the value of last element is approximately half that of the sum-- and if that uncountable-- or we could sum up the values, as there were a countable sequence, the first element 1, the second 2, etc, and get the "average value" or expected value, and if that is uncountable which must be if the sum is uncountable (i presume) because the denominator is only countably large. Also in what sense are there an uncountable of paths, which i presume is how storrs mccalls infers that its as if there are uncountably many leaves. The number of paths seem to grow via the same limiting process 2,4,8,16, and at no point does it every stop or actually get to infinity if you suggest it somehow stopps then isnt that aki to treating the infinity as an actual entity, as in nonstandard analysis (by the way i storrs mccall may be using in his model)
OK. One more time. When I counted the branchings, I looked solely at the configuration
o
/ \
for each node n the tree.There are 2^n - 1 of these in a tree with n levels.
Paths in the finite tree are paths from the root node to the leaves. A finite tree with n levels has 2^n leaves and 2^n paths to those leaves. However, when you take limits and imagine a tree of infinite depth, then *there are no more leaves*. A leaf is a terminal node, and the infinite tree by definition does not have them.
Your problem still is that you look at the finite trees, count the leaves and find that there are 2^n of those at level n (if you count the root node as living at level 0). You then go, formally or informally,
Let K_n be the set of paths to nodes at the n'th level. Its cardinality, card(K_n) is 2^n, so
lim n->oo card(L_n) = "countable".
Never mind that you haven't defined the limit precisely (although this could be done) and that the (classical) limit of the sequence {1, 2, 4, 8, ...} does not exist.
Your major mistake, however, is that you assume that you can somehow define K_oo as a limit of the trees with n levels and that then the cardinality of the paths in the infinite tree is
card(K_oo) = card(lim i->oo K_i) (so far so good) = (????) lim(i->oo card(K_i).
The second step in this equality is unjustified and unjustifiable. On the other hand, there are clear proofs that the set of paths in the infinite tree is uncountable, i.e.,
card(lim i->oo K_i) > lim(i->oo card(K_i).
well i mean i think this what i was getting in getting the cardinality of the number of paths; one is not taking a limit. ONe appear to be taking it as a totality.
When you say card(lim i->oo K_i)> lim(i->oo card(K_i). It appears that on the rhs, you are treating the infinity concept as a totality and not a limit; as an actual infinite number that exists. That was why i mentioned a nonstandard analysis approach to the RHS which treats these limits, not as limits but as actual totalities as well. It appears to be an artefact of the mathematical system.
Suppose i had the sequence, 2, 4,8, 16 etc and i denote the first element A1 (ie 2), A2 for the second Ie (4) and i then to say card (lim n_>infinity ki)--- where it assumed that this sequence is not just an observed sequence but a function of underlying mechanism, rule such that an=2^N so that we looking at the process and not the product. For example in the case of st petersburg paradox where it is explicitly defined that the utility for n outcome is (2^N) and its probability 1/(2^N)- I am not how they compute this limit to get the expected value (off the top of my head to be infinite-- presumably it just results from a constant sum mitigated by the inverse probabilities but if one were to look at the probabilities and utilities individual they are unbounded and could not fit into an infinite sequence (within a frquentist paradigm let say)- especially if one were to take what is called a hyper hyperthetical frequentist approach which uses nonstandard analysis
You say: "When you say card(lim i->oo K_i)> lim(i->oo card(K_i). It appears that on the rhs, you are treating the infinity concept as a totality and not a limit;"
Not so. K_i is a finite set, and it has a well-defined cardinality, namely 2^n. I can ask about the limit of this cardinality (which is a number) as i->oo. Nothing unusual about that. It is not even necessary for that limit to exist for you to accept that
card(lim i->oo K_i) =/= lim(i->oo card(K_i).
The main point is that you yourself act as though the infinite tree is a closed entity, i.e., you accept that lim i->oo K_i exists in some sense. But then you can ask about its cardinality, which is card (lim i->oo K_i) and not lim(i->oo card(K_i). There is no ambiguity.
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