Re: Cantor and the binary tree

Jan de Vos said:
> In sci.math, Tony Orlow wrote:
> >> This, I regret to inform you, is not true. But you can try to prove
> >> me otherwise by giving a mapping from the set of paths into the set of
> >> branches.
> >>
> > Been there, done that. Every 0 branch is a continuation of the
> > parent's path, with the same quantitative value. Every 1 branch
> > initiates a new path with a new value. So, there are half as many
> > paths as branches. I don't understand why this is so hard for you to
> > see. Look at finite examples, and see if anything changes as you
> > approach infinity. It doesn't.
>
> This way, you only count the kinds of path that go down '0' branches
> from some point on, not paths that do something else, like repeat '0'
> and '1' indefinitely, or do something completely random.
>
> And I didn't ask for some handwaving with a conclusion about how many
> things there are, I expressely ask for a mapping. In other words,
> give me a way to tell, for /any/ path, to which node it corresponds.
>
> The thing we disagree on (methinks), is that you think every path ends
> in a leafnode, which might be a node infinitely low. Everyone else
> here, though, don't believe such nodes exist -- iow, each node is at
> some finite height in the tree. Paths, on the other hand, do go on
> forever, just like the natural numbers. So perhaps it is more logical
> to can the whole path-counting exercise until we agree on that.
Um, you want me to agree that you have an infinite binary tree with only finite
paths, or infinite paths with only finite numbers of nodes, or infinite numbers
of nodes a finite distance from the root. I'm sorry but those ideas make no
sense, and I can't agree to any of that.
>
>
> Jan
>
>

--
Smiles,

Tony
.

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