08 Sep 04
Originally posted by iamatigerOK, I asked the wrong question, as 'generated' takes care of all that. I have no doubt that such an extension exists (let's call it X), has some nice properties and has uncountably many elements. The quesion should have been, what properties does X have?
No, there have to be an infinite number of different infinitessimals for it all to work. If b is an infitessimal number then 2.b is a larger infinitessimal number.
Here are two: as a subset of X, R is bounded by +-1/d and has the discrete topolgy.
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08 Sep 04
Originally posted by AcolyteHere's a good one - the set of all infinitessimals (call it x) has no upper or lower bound.
OK, I asked the wrong question, as 'generated' takes care of all that. I have no doubt that such an extension exists (let's call it X), has some nice properties and has uncountably many elements. The quesion should have been, what properties does X have?
Here are two: as a subset of X, R is bounded by +-1/d and has the discrete topolgy.
If x's upper bound was some infinitessimal a, then 2a would also be an infinitessimal, which then is a larger potential upper bound. If x's upper bound was some real number r then r/2 would be a smaller real number which is a smaller potential upper bound. Same arguments hold for the lower bound. Neither bound can be infinity (as for the set of reals) because all infinitessimals are much smaller than 1)
09 Sep 04
Originally posted by iamatigerA set doesn't have to have a 'nearest bound' to be bounded. For example, the set of all rationals with squares less than 2 is bounded, but it has no nearest bounds in Q.
Here's a good one - the set of all infinitessimals (call it x) has no upper or lower bound.
If x's upper bound was some infinitessimal a, then 2a would also be an infinitessimal, which then is a larger potential upper bound. If x's upper bound was some real number r then r/2 would be a smaller real number which is a smaller potential upper bound. Sam ...[text shortened]... d can be infinity (as for the set of reals) because all infinitessimals are much smaller than 1)
x is bounded by +-r, where r is any real number.
Given a in x, (a-d,a+d) is also in x, so x is open. (I'm counting 0 as part of x for convenience).
We can express all non-infinitesimals as r + b, where r is real nonzero and b is in x. In this case (r,r+2b) is also in X\x. So X\x is open.
=> X is disconnected. Is there any way of extending it further to fill in the gaps?
09 Sep 04
Originally posted by rgoudieI don't see why not - all that article says is that infinitesimals don't obey the least upper bound axiom, which is pretty obvious: the LUB axiom is equivalent to saying 1 - 0.999... = 0.
One cannot really manipulate infinitesimals like that. 🙂
http://en.wikipedia.org/wiki/Infinitesimal
-Ray.
The set of infinitesimals may not have nearest bounds, but here's something nice about it: it's a ring-without-1, a vector space over R, and a module over the finites. The set of infinities is not even closed under addition.
It looks like the simplest way of actually describing an element of X is to represent it by real numbers x(i), where i is an integer, so that X = Sum x(i)d^i . In this case, if we ignore multiplication the 'layers' of X are pretty similar to each other: the finites are to X as the infinitesimals are to the finites, and so on.
09 Sep 04
2 edits
Originally posted by iamatigerThe reals don't have nearest bounds either - why take infinity when you could have infinity/2?
Here's a good one - the set of all infinitessimals (call it x) has no upper or lower bound.
If x's upper bound was some infinitessimal a, then 2a would also be an infinitessimal, which then is a larger potential upper bound. If x's ...[text shortened]... set of reals) because all infinitessimals are much smaller than 1)
09 Sep 04
It's equal to delta (which is used in Calculus to represent any small number so close to zero, it's just represented by this).
Or, you might want to look into Newton's laws of "limits" as x approaches zero. It's the foundation for calculus and is very simple. I don't know if someone has already said any of this b/c I didn't want to read through every post. So I offer my potential appologies.
09 Sep 04
1 edit
Originally posted by fiendizeNewton thought there was something suspect about his calculus, which is why he avoided it in Principia. Modern standard analysis, which was the first successful attempt to deal with calculus in a rigorous way, did not emerge until the 19th century. So while Newton can reasonably claim to have invented calculus, his system has since been replaced by a better one, much like his laws of forces have been replaced by relativity.
It's equal to delta (which is used in Calculus to represent any small number so close to zero, it's just represented by this).
Or, you might want to look into Newton's laws of "limits" as x approaches zero. It's the foundation fo ...[text shortened]... nt to read through every post. So I offer my potential appologies.
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14 Sep 04
Originally posted by Acolyteas infinity/2 = infinity, the reals are bounded by +- infinity. What I mean with the infinitessimals is you cannot compute or name any number at all which is >= all infinitessimals and <= all non infinitessimals.
The reals don't have nearest bounds either - why take infinity when you could have infinity/2?
14 Sep 04
Originally posted by iamatigerEr, not if we want X to be a group under multiplication. Just as d and 2d are distinct infinitesimal numbers, 1/d and 1/2d are distinct infinite numbers. You're right about the gap between infinitesimals and non-infinitesimals though - X is peppered with holes.
as infinity/2 = infinity, the reals are bounded by +- infinity. What I mean with the infinitessimals is you cannot compute or name any number at all which is >= all infinitessimals and <= all non infinitessimals.
16 Sep 04
I am still a little boggled by this whole discussion. Chicken said .9999... = 1, so I will agree, but I need this explained to me. I see it as.....if you asked yourself if two #'s are the same, their diff. is zero, so
2 - 2 = 0, but 1-.9999.. != 0, so I dunno really. Someone please refute this!
16 Sep 04
Originally posted by !~TONY~!It's the guy above you, and not me, to whom you should look for guidance.
I am still a little boggled by this whole discussion. Chicken said .9999... = 1, so I will agree, but I need this explained to me. I see it as.....if you asked yourself if two #'s are the same, their diff. is zero, so
2 - 2 = 0, but 1-.9999.. != 0, so I dunno really. Someone please refute this!
My proofs that 0.9 rec = 1 are sound in standard analysis, although the very basic level of that is something I know very little about. However, while I haven't read it carefully, there seem to be changes in the axioms governing the real numbers that leave most of analysis intact but allow for infinitesimals.
16 Sep 04
Originally posted by royalchickenRight....your proof was pretty convincing...but the simplicity of mine is way more impressive! (Note sarcasm) I got it now....
It's the guy above you, and not me, to whom you should look for guidance.
My proofs that 0.9 rec = 1 are sound in standard analysis, although the very basic level of that is something I know very little about. However, while I haven't read it carefully, there seem to be changes in the axioms governing the real numbers that leave most of analysis intact but allow for infinitesimals.