Can "Infinity" ever be more than a mathematical abstraction?

[arfa brane]

You never seem to have a point or a thesis.

Thanks. I'm trying really hard not to.

You communicate like a bot.

And you communicate like a disparaging, condescending twat. Mostly.

Y0u come across like someone who thinks they have all the right ideas. That's probably just youthful arrogance. Please don't get the idea I'm trying to give myself some comfort here, I don't really care

 

[someguy1]

Thanks. I'm trying really hard not to.And you communicate like a disparaging, condescending twat.

Then why do you keep replying to my posts? I'd be perfectly happy if you stopped.

You know for several pages you've been throwing out random math phrases you've read on Wikipedia and asking me to put them in context for you. Why are you doing this? To troll me? Guess you got me good.

 

[arfa brane]

Then why do you keep replying to my posts? I'd be perfectly happy if you stopped.

So would I. But this thread isn't about you or me, is it?

 

[someguy1]

So would I. But this thread isn't about you or me, is it?

Hey man I'm sorry if I upset you. Or anyone. You keep asking these questions and I'm answering the best I can but I can't tell where they're coming from. At some point I lost patience. My apologies. Time for a forum break. Peace.

 

[arfa brane]

You know for several pages you've been throwing out random math phrases you've read on Wikipedia and asking me to put them in context for you.

The quotes aren't from Wikipedia and I haven't had the least expectation that you will put them in context for me. Nor have I asked you or anyone else to do that. Your arrogance is a given, it seems. You can't help yourself.

To the rest:

The ideas I've been trying to explore go like this: the real line is continuous, so is physical motion. With calculus you can define a set of instantaneous positions and velocities for a classical object with motion. Calculus usually defines the derivative in terms of limits, the epsilon-delta method uses real numbers, not infinitesimals (mainly because there's only one infinitesimal real). But the derivative can be defined in other ways; if you want to invoke infinitesimals you "need" a map from them to the standard reals.

Also, when you see a textbook saying dy/dx means an infinitesimal change in y and in x in the reals, they don't mean actual infinitesimals but very small real numbers (one supposes).


What calculus seems to say is, it doesn't matter if the real line is after all a subset of the hyperreal line, or if we call one or the other an abstraction of reality. Zeno's problems with continuity of motion seem to have more than one solution. And Zeno was just saying there's a problem if we assume real (actual) distances are infinitely divisible. (that is, into 1/2 the distance, then 1/2 the remaining distance, ad infinitum), so he thought, an object in motion has to pass through an infinite number of "checkpoints" in a finite time. He got that wrong, but why?

Is it because the real line isn't, after all, infinitely divisible? Or is it because an object's motion really isn't "from point to point"?

 

[iceaura]

Come on, man. You contradicted established mathematical fact. - - .

I did not. I singled out and removed from the discussion a difficulty, or confusion, that results from using inappropriate physical metaphors - such as "holes" - to refer to mathematical objects and features.
There are no physical holes, or anything corresponding to physical holes, in the computable reals. There are infinities - limits, such as the limit of the series 1+1/2+1/4 - - -. It is equal to "2".

Calculus doesn't claim anything exists.

In particular, it does not claim that the misleadingly labeled noncomputable "holes" in the computable real number line correspond to, abstract, any physical reality whatsoever.
And as far as we have any evidence, they don't.

I noted earlier that the rationals model the real world too, to any desired degree of precision. You (or someone) disagreed and said that we need all the computables for some reason or other. That's not so, you can arbitrarily approximate any computable number with rationals.

If you don't have the number, you don't have what you are approximating. Your model is missing - say - the diameter of a circle. If you don't need the diameter of a circle to model your physical reality, then ok - you don't need pi. If you do need the diameter of a circle, you do need pi.

 

[TheFrogger]

Not neccessarily. The diameter is one centremetre. Pi is not needed.

 

[TheFrogger]

Incidentally if every number could equal the same number, does it continue for infinity?

n-n=0

In this way ANY number can equal zero.

 

[Write4U]

Incidentally if every number could equal the same number, does it continue for infinity?

n-n=0

In this way ANY number can equal zero.

hehe...I see it that (n1) can be any number and be approximate to (n2) with a difference of zero, between the measurement of each.

 

[TheFrogger]

So mathematics should be a mess! The Laws should not exist in this case!

 

[Michael 345]

Is it because the real line isn't, after all, infinitely divisible? Or is it because an object's motion really isn't "from point to point"?

Is it because the real line isn't, after all, infinitely divisible?

In THEORY yes
Practice NO

I would contend Planck length would be a answer your looking for

http://www.physlink.com/education/askexperts/ae281.cfm

Or is it because an object's motion really isn't "from point to point"?

If you are thinking about the runner not passing the tortoise

In THEORY yes but only if you restrict the distance to a finite length

If you restrict the distance, as a example, to 100 metres, again in THEORY, runner does not pass tortoise (see also Planck length)

Allow unlimited distance tortoise looses in a very short period or distance

 

[QuarkHead]

Is it because the real line isn't, after all, infinitely divisible?

In THEORY yes
Practice NO

I don't think this quite right. Of course, I may have misunderstood this exchange, but try this.

Take the Real Line $$R^1$$ (this btw is a topological construct). Assume for a contradiction that it can be arbitrarily partitioned into 2 disjoint sets, $$L$$ and $$U$$ such that $$L\cap U = \emptyset$$.

Now every element in $$L$$ is a lower bound for every element in $$U$$, and every element in $$U$$ is an upper bound for every element in $$L$$. But since $$R^1$$ is equipped with a a total order, there is a greatest element in $$L$$ which is the greatest upper bound for $$L$$ and therefore in $$U$$.

Likewise there is a least element in $$U$$ is the least lower bound for $$U$$, and therefore is in $$L$$.

Therefore my assumption fails, this partition into disjoint sets cannot exist, which implies that the Real line is "continuous" in the colloquial sense of the word

t

 

[iceaura]

Not neccessarily. The diameter is one centremetre. Pi is not needed.

Without pi, you have no circle to go with your linear distance - your one centimeter is not a diameter.

Now every element in LLL is a lower bound for every element in UUU, and every element in UUU is an upper bound for every element in LLL. But since R1R1R^1 is equipped with a a total order, there is a greatest element in LLL which is the greatest upper bound for LLL and therefore in UUU.

Likewise there is a least element in UUU is the least lower bound for UUU, and therefore is in LLL.

The condition that every element of one is a bound for the other does not imply that every bound for the one is an element of the other.

 

[QuarkHead]

The condition that every element of one is a bound for the other does not imply that every bound for the one is an element of the other.

Do you not understand the process of "proof by contradiction"?

 

[iceaura]

Do you not understand the process of "proof by contradiction"?

Each appearance of the word "therefore", in your proof, is an error.

 

[QuarkHead]

Each appearance of the word "therefore", in your proof, is an error.

As I said earlier, you do not understand proof by contradiction. My "therefores" where part of the assumption I proved wrong. That is how this sort of proof works.

 

[someguy1]

Take the Real Line $$R^1$$ (this btw is a topological construct). Assume for a contradiction that it can be arbitrarily partitioned into 2 disjoint sets, $$L$$ and $$U$$ such that $$L\cap U = \emptyset$$.

Like $$L = \mathbb Q$$ and $$U = \mathbb R \setminus \mathbb Q$$? Perhaps you left out a vital assumption.

Now every element in $$L$$ is a lower bound for every element in $$U$$,

Surely false with the example I just gave. Methinks you forgot to say:

... And every element of $$L$$ is strictly less than every element of $$R$$. Right? That's the part you left out.

So, let $$L = (-\infty, 0]$$ and $$R = (0, +\infty)$$. That satisfies your hypotheses, including the one you forgot to mention.

and every element in $$U$$ is an upper bound for every element in $$L$$.

True.

But since $$R^1$$ is equipped with a a total order, there is a greatest element in $$L$$ which is the greatest upper bound for $$L$$ and therefore in $$U$$.

$$0$$ is indeed the greatest element in $$L$$, but it's not in $$U$$.

Likewise there is a least element in $$U$$ is the least lower bound for $$U$$, and therefore is in $$L$$.

No, there is no least element in $$L$$. You're wrong here.

Therefore my assumption fails, this partition into disjoint sets cannot exist, which implies that the Real line is "continuous" in the colloquial sense of the word

It's your "proof" that fails, since even AFTER stating the assumption you failed to mention, the most obvious example falsifies it.

Perhaps you forgot to mention that the sets are assumed to be open, in which case you could work out a proof (but definitely not this one) that the reals are connected. Not "continuous," which has no meaning in this context.

@QH, How many unstated assumptions are the readers supposed to supply? Even after assuming all your missing assumptions, the proof's still wrong. That is, if the readers figure out that (1) everything in L is less than everything in R; and (2) L and R are open sets; then you STILL haven't given a proof that the reals are connected.

 

[iceaura]

My "therefores" where part of the assumption I proved wrong.

They are mistakes in deduction.

The condition that every element of one is a bound for the other does not imply that every bound for the one is an element of the other.

 

[QuarkHead]

No, there is no least element in $$L$$. You're wrong here.

I challenge you to find where I ever claimed that

@QH, How many unstated assumptions are the readers supposed to supply?

Then you are being disingenuous. I made 2 assumptions - that not many here have a grounding in topology, and that nonetheless they would be happy with the intuitive notion that the Real Line is simply all the Real numbers arranged in the usual order as some sort of "line".

The "partition" I described, as you will know very well, is called a Dedekind Cut

 

[iceaura]

The "partition" I described, as you will know very well, is called a Dedekind Cut

Someguy mentioned a problem with your description, I mentioned a problem with your reasoning, here's a closer look:

Now every element in LL is a lower bound for every element in UU, and every element in UU is an upper bound for every element in LL.

Let's accept that under advisement, despite someguy's correct objection (it's intuitively ok, once past the linguistic hitch of bounded elements)

But since R1R1 is equipped with a a total order, there is a greatest element in LL {1} which is the greatest upper bound for LL {2} and therefore in UU {3} .

1} is false as it stands - there are many such sets LL that have no greatest element.
2} is false - every element of UU is an upper bound of L, and UU has no greatest element - and does not make sense in your argument, where you seem to need a least upper bound for LL.
3} is false as it stands, since many sets LL as described (the ones that have a greatest element) do contain their {least} upper bound - it isn't in UU.