Re: An uncountable countable set
- From: Tony Orlow <tony@xxxxxxxxxxxxx>
- Date: Sun, 01 Oct 2006 15:14:21 -0400
imaginatorium@xxxxxxxxxxxxx wrote:
Tony Orlow wrote:imaginatorium@xxxxxxxxxxxxx wrote:Randy Poe wrote:That all makes very good sense, Brian. I can't see that there isn't aTony Orlow wrote:No, but I think the problem is elsewhere, slightly. Is there a formalVirgil wrote:If something doesn't occur, the question "when does it occur"In article <451bac34@xxxxxxxxxxxxxxxxxxx>,So, you have no answer.
Tony Orlow <tony@xxxxxxxxxxxxx> wrote:
It does answer the question of "whether" it occurs. "When" is of lesserDoes that answer the question of **when** this occurs? Of course not.If the vase is empty at noon, but not before, how can that not be theSaying that it is empty is quite different from saying anything about a
moment that it becomes empty?
"last ball". andy does not deny that the vase becomes empty, he just
does not say anything about any "last ball out".
importance.
does not have an answer.
definition of what "transition" means? (Not in a nearby pocket "Dict.
of maths." for example)
Seems to me that if you had the graph y = (1 if x<0; 2 if x>=0), and
and associated state transition diagram, then there would be a
"transition" from 1 to 2 "at" x=0. But such terminology does not
capture what the state is "at the point of" the transition, which may
be why it isn't used much. But if a vase has balls in it for values (-1
<= t < 0), I can't see anything actually wrong with saying there is a
transition from empty to non-empty "at" t=-1 and a transition from
non-empty to empty "at" t=0. You need to be careful not to deduce
anything about the _state_ at the two transition values.
After all, consider the graph of y = -x. This is positive for x<0, and
negative for x>0. It's undefined for x=0, but is there not a transition
from positive to negative at x=0?
point of transition, especially when that point is very explicitly
defined to be noon, and that it is at least then, and not until then,
that the vase goes from being non-empty to empty.
Yes, but using this (slightly loose?) "transition" terminology, you
have to be careful that this tells you _nothing_ about the state _at_
noon. Seems to me that all of the following graphs have a "transition"
from positive to negative at x=0, but very different things are true
_at_ 0
"_nothing_"?
y=-1/x (at x=0 this is undefined)
Or y=1/x. From one side it goes to oo, and from the other, -oo. This is resolved by application of the number circle concept, where oo=-oo. Then, while the value is "undefined" in the sense of being infinite, it's at least consistent with itself.
y = (1 if x<0; -1 if x>=0) : y(0) = -1Clearly discontinuous - two different formulas
y = (1 if x<=0; -1 if x>0) : y(0) = 1ditto
f = 1 if x<0; -1 if x>0; purple unicorn if x=0 : well, what can I say?That's enough. None of those examples pertains to the vase, which is a very well defined increment of 9 per iteration. To put it in limit terms, the "solution" to the vase problem would be equivalent to saying lim(x->oo: 9x)=0. Ain't the case.
Is y=-x really "undefined" at x=0? I don't think so. It's 0, which
equals -0.
Sorry, that's a (surely obvious?) typo - I meant y=-1/x. No ok, perhaps
it isn't obvious.
Okay, but oo can be viewed as equivalent to -oo, in some contexts.
But another example:
y=-x : y(0) = 0, which is perfectly well defined, but neither positive
nor negative (au moins en anglais)
Yes, which fits perfectly well with 0's reciprocal, oo, or 100...000, which is also its own additive inverse.
(I'll carry on with the blue sliver when I have time, which may be in a
day or so)
Brian Chandler
http://imaginatorium.org
Carry on!
T
.
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