Working with the set of real numbers
Rotwang
sg552 at hotmail.co.uk
Thu Feb 13 23:21:40 CET 2014
On 13/02/2014 22:00, Marko Rauhamaa wrote:
> Rotwang <sg552 at hotmail.co.uk>:
>
>>> for x in continuum(0, max(1, y)):
>>> # Note: x is not traversed in the < order but some other
>>> # well-ordering, which has been proved to exist.
>>> if x * x == y:
>>> return x
>>
>> [...]
Restoring for context:
>>> The function could well return in finite time with a precise result
>>> for any given nonnegative real argument.
>> More importantly, though, such a computer could not complete the above
>> iteration in finite time unless time itself is not real-valued. That's
>> because if k is an uncountable ordinal then there is no strictly
>> order-preserving function from k to the unit interval [0, 1].
>
> If you read the code comment above, the transfinite iterator yields the
> whole continuum, not in the < order (which is impossible), but in some
> other well-ordering (which is known to exist). Thus, we can exhaust the
> continuum in ℵ₁ discrete steps.
Yes, I understood that. But my point was that it can't carry out those
ℵ₁ discrete steps in finite time (assuming that time is real-valued),
because there's no way to embed them in any time interval without
changing their order. Note that this is different to the case of
iterating over a countable set, since the unit interval does have
countable well-ordered subsets.
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