Insights Why Do People Say That 1 and .999 Are Equal? - Comments

[Multiple_Authors]

Multiple_Authors submitted a new PF Insights post

Why Do People Say That 1 And .999 Are Equal?

Continue reading the Original PF Insights Post.

 

[HallsofIvy]

While there are several arguments here that people might accept as proofs, they all, as stated, are non-rigorous, involving some basic, unstated assumptions. For example, in argument two, it is assumed that multiplying x= 0.999... by 10 gives 10x= 9.999... and then that subtracting x give 9x= 0.999... again. Both of those assume the usual arithmetic properties are true for 0.999... That is true but exactly the sort of thing people who object to "0.999...= 1" would object to anyway. In argument four, it is accepted that 0.333...= 1/3. Why would a person who objects to "0.999...= 1" accept that?

It is not too difficult to give a rigorous proof using "geometric series". It is easily shown that the sum \sum_{i= 0}^n ar^i is \frac{a(1- r^n)}{1- r}. To do that, let S_n= \sum_{i= 0}^n ar^i= a+ ar+ ar^2+ \cdot\cdot\cdot+ ar^n). Since that is a finite sum, the usual properties of arithmetic hold and we can write S_n- a= ar+ ar^2+ \cdot\cdot\cdot+ ar^n= r(a+ ar+ \cdot\cdot\cdot+ ar^{n-1}). The quantity in the last parentheses is almost S_n itself. It is only missing the last term, ar^n. Restore that by adding ar^{n+1} to both sides: S_n- a+ ar^n= r(a+ ar+ \cdot\cdot\cdot+ ar^{n-1})+ ar^{n+ 1}. Taking that last term inside the parentheses we have S_n- a+ ar^n= r(a+ ar+ \cdot\cdot\cdot+ ar^{n-1}+ ar^n)= rS_n. We can write that as rS_n- S_n= (r- 1)S)_n= ar^n- a= a(r^n- 1) and dividing both sides by r- 1, S_n= \frac{a(r^n- 1)}{r- 1} which, for r< 1 can be written more as S_n= \frac{a(1- r^n)}{1- r}

The geometric series, \sum_{i= 0}^\infty ar^i, is, by definition, the limit of the "partial sums", S_n= \sum_{i= 0}^n ar^i, as n goes to infinity. As long as |r|&lt; 1, \lim_{n\to\infty} r^n= 0 so that limit is easily calculated as \sum_{i= 0}^\infty ar^i= \frac{a}{1- r}.

Now, turning to the problem at hand. 0.999... is, by definition of the decimal numeration system, 0.9+ 0.09+ 0.009+ \cdot\cdot\cdot= 0.9+ 0.9(0.1)+ 0.9(0.01)+ \cdot\cdot\cdot= 0.9+ 0.9(.1)+ 0.9(.1^2)+ \cdot\cdot\cdot, precisely a geometric series with a= 0.9 and r= 0.1. So the sum, and the value of 0.9999..., is \frac{0.9}{1- 0.1}= \frac{0.9}{0.9}= 1.

 

[Greg Bernhardt]

While there are several arguments here that people might accept as proofs, they all, as stated, are non-rigorous, involving some basic, unstated assumptions.

Here is the previous FAQ
https://www.physicsforums.com/insights/is-there-a-rigorous-proof-of-1-0-999/

 

[Algr]

Proofs #2 & #3 are what convinced me that 1 and .999... are NOT equal. It's just insulting to assume that .333... = 1/3 in this context. And why assume that .999... and 9.999... have different numbers of nines when this would never happen for any finite number?

 

[pabilbado]

I alread covered this in blog a while ago, though it's in spanish: https://preguntaspococientificas.wordpress.com

 

[FactChecker]

If they are different, how different are they? Suppose you say that they are different by more than 0.0001. Then 1 - 0.99999 is closer and the more 9's you add, the closer it gets. So you can not say there is any difference.

 

[SlowThinker]

If they are different, how different are they? Suppose you say that they are different by more than 0.0001. Then 1 - 0.99999 is closer and the more 9's you add, the closer it gets. So you can not say there is any difference.

I'm not an expert on these things, but you can easily tell that a proof is wrong when it proves 0.9999...=1 on surreal numbers. Which most of the above do.
I'm not entirely sure about HallsofIvy's proof, it might be OK, even if I think it's swapping $\infty$ with $\omega$ somewhere along the way.

 

[FactChecker]

I'm not an expert on these things, but you can easily tell that a proof is wrong when it proves 0.9999...=1 on surreal numbers. Which most of the above do.
I'm not entirely sure about HallsofIvy's proof, it might be OK, even if I think it's swapping $\infty$ with $\omega$ somewhere along the way.

You can easily prove it by contradiction, as I indicated. Once that is done, you know that they are equal. Then I would to be careful about saying that other proofs are wrong unless I completely understood the other proofs and can pinpoint the error.

 

[Mark44]

Proofs #2 & #3 are what convinced me that 1 and .999... are NOT equal. It's just insulting to assume that .333... = 1/3 in this context.

How is this assumption insulting. Just by ordinary division you can show that the 3 digit keeps repeating.

And why assume that .999... and 9.999... have different numbers of nines when this would never happen for any finite number?

To the right of the decimal point, both have the same number of 9 digits. Also I don't understand what you are saying about "finite numbers". Both .999... and 9.999... are finite numbers; i.e., strictly less than infinity.

I'm not an expert on these things, but you can easily tell that a proof is wrong when it proves 0.9999...=1 on surreal numbers. Which most of the above do.

I did a quick scan of all of the proofs. None of them mentioned surreal numbers. If you can point out where I might have missed this, I would appreciate it.

I'm not entirely sure about HallsofIvy's proof, it might be OK, even if I think it's swapping $\infty$ with $\omega$ somewhere along the way.

Please show me where.

 

[SlowThinker]

I did a quick scan of all of the proofs. None of them mentioned surreal numbers. If you can point out where I might have missed this, I would appreciate it.

My idea was: if the proof, applied over surreal numbers, proves that 0.9999...=1, then the proof is not correct. There must be some implicit assumption.

Please show me where.

Now, turning to the problem at hand. 0.999... is, by definition of the decimal numeration system, $0.9+ 0.09+ 0.009+ \cdot\cdot\cdot$ = $0.9+ 0.9(0.1)+ 0.9(0.01)+ \cdot\cdot\cdot$ = $0.9+ 0.9(.1)+ 0.9(.1^2)+ \cdot\cdot\cdot$, precisely a geometric series with a= 0.9 and r= 0.1. So the sum, and the value of 0.9999..., is \frac{0.9}{1- 0.1}= \frac{0.9}{0.9}= 1.

I would say that, by definition of the decimal numeration system, 0.9999...=$\sum_{n=1}^\omega{}9\ 0.1^n$ rather than $\sum_{n=1}^\infty{}9\ 0.1^n$. The difference is, obviously, 0 in real numbers.

 

[Mark44]

I did a quick scan of all of the proofs. None of them mentioned surreal numbers. If you can point out where I might have missed this, I would appreciate it.

My idea was: if the proof, applied over surreal numbers, proves that 0.9999...=1, then the proof is not correct. There must be some implicit assumption.

Why would the proof need to invoke surreal numbers? The proofs are all concerned with two representations of the same real number.

Please show me where.

Now, turning to the problem at hand. 0.999... is, by definition of the decimal numeration system, $0.9+ 0.09+ 0.009+ \cdot\cdot\cdot$ = $0.9+ 0.9(0.1)+ 0.9(0.01)+ \cdot\cdot\cdot$ = $0.9+ 0.9(.1)+ 0.9(.1^2)+ \cdot\cdot\cdot$, precisely a geometric series with a= 0.9 and r= 0.1. So the sum, and the value of 0.9999..., is \frac{0.9}{1- 0.1}= \frac{0.9}{0.9}= 1.

I would say that, by definition of the decimal numeration system, 0.9999...=$\sum_{n=1}^\omega{}9\ 0.1^n$ rather than $\sum_{n=1}^\infty{}9\ 0.1^n$. The difference is, obviously, 0 in real numbers.

And that's the set we're concerned with here.

 

[SlowThinker]

Why would the proof need to invoke surreal numbers? The proofs are all concerned with two representations of the same real number.

If the proof can be used to prove a false statement, then the proof must be wrong.
0.9999...$\neq$1 in surreal numbers, so if the proof works with them too, then it's wrong.
I dare say that all of the "informal proofs" from the article immediately work with surreal numbers, so they must be wrong.

This quote from the original article,

Can we define number systems such that 1=0.999… does not hold? Of course!

seems to be the answer that people are looking for, together with "Real numbers are defined in such a way that 0.999...=1".

 

[Ernest S Walton]

So does this mean that .99999... = .99999...8 ?

 

[pwsnafu]

So does this mean that .99999... = .99999...8 ?

No, because the right hand side is nonsense. You can't have a non-terminating run of nines and then terminate it.

 

[Ernest S Walton]

No, because the right hand side is nonsense. You can't have a non-terminating run of nines and then terminate it.

Okay... but it seems like to make .999 = 1 you also have to terminate the run of nines. Anyway I know this has been discussed several times so I'll read up some more.

 

[Mark44]

So does this mean that .99999... = .99999...8 ?

As already mentioned, the right side is meaningless for two reasons. On the left side the ellipsis (...) means that the pattern you see repeats endlessly. On the right side, it is not known how many 9s there are in front of the final 8 digit.

Okay... but it seems like to make .999 = 1 you also have to terminate the run of nines.

Don't confuse .999 with .999... They mean two very different things. The first is identical to 999/1000 which is smaller than 1, and the second is equal to 1.

Anyway I know this has been discussed several times so I'll read up some more.

 

[Nemika]

Proofs #2 & #3 are what convinced me that 1 and .999... are NOT equal. It's just insulting to assume that .333... = 1/3 in this context.

How is this assumption insulting. Just by ordinary division you can show that the 3 digit keeps repeating.

And why assume that .999... and 9.999... have different numbers of nines when this would never happen for any finite number?

To the right of the decimal point, both have the same number of 9 digits. Also I don't understand what you are saying about "finite numbers". Both .999... and 9.999... are finite numbers; i.e., strictly less than infinity.

I'm not an expert on these things, but you can easily tell that a proof is wrong when it proves 0.9999...=1 on surreal numbers. Which most of the above do.

I did a quick scan of all of the proofs. None of them mentioned surreal numbers. If you can point out where I might have missed this, I would appreciate it.

I'm not entirely sure about HallsofIvy's proof, it might be OK, even if I think it's swapping $infty$ with $omega$ somewhere along the way.

Please show me where.

*Both .999… and 9.999… are finite numbers; i.e., strictly less than infinity.*

How can you call 0.999... or 9.999... as finite?
If these are finite then infinity can also be considered finite!

 

[WWGD]

*Both .999… and 9.999… are finite numbers; i.e., strictly less than infinity.*

How can you call 0.999... or 9.999... as finite?
If these are finite then infinity can also be considered finite!

Please look up the definition of a finite number. There is a difference between an infinite _decimal expansion_ and an infinite number.

 

[Mark44]

*Both .999… and 9.999… are finite numbers; i.e., strictly less than infinity.*

How can you call 0.999... or 9.999... as finite?
If these are finite then infinity can also be considered finite!

Please look up the definition of a finite number. There is a difference between an infinite _decimal expansion_ and an infinite number.

WWGD is correct. Finite numbers are bounded; infinite numbers are unbounded.

 

[mathexam]

They are clearly different but if you're approximating, they're almost equal.

 

[Mark44]

They are clearly different but if you're approximating, they're almost equal.

Who are "they"? Are you referring to .999 and .999... ?

Clearly these two numbers are close, but that's not the point. What we're saying is that .999... is identically equal to 1, and .999 is smaller than 1 by .001.

 

[Isaac0427]

I have no way to say this is true for sure, but it appears to me as though proof #3 is based off an incorrect assumption. Wouldn't the difference between .999... and 1 be infinitesimal, not 0. Isn't this one of the reasons we have the concept of infinity, and in this case an infinitesimal value. If something infinitesimal is just considered 0, wouldn't a Riemann sum not mater at all, and just be equal to zero? Wouldn't that make anything using dx, dy, etc. represent a change of zero, as they are defined as an infinitesimaly small change in their respective variable? This is just my thinking, I am not being ignorent, I just want to understand.

 

[SlowThinker]

Wouldn't the difference between .999... and 1 be infinitesimal, not 0.
something infinitesimal is just considered 0

Both are true. The important part is that real numbers are not precise enough to distinguish an infinitesimal difference - which is, in fact, the reason why we can use infinitesimals to integrate and derive expressions containing real numbers.

I would think this whole theme is answered by:
The definition of real numbers says that $$0.999..._{\text{as real number}} \ =_{\text{acting on real numbers}} \ 1_{\text{as real number}}$$
If you don't like it, use other numbers.

 

[madness]

I think equivalence is more intuitive if you simply state that the limit of 0.999... as we keep adding more decimal places is 1. I think this is also a closer translation of the mathematical statement into plain english.

 

[Isaac0427]

I think equivalence is more intuitive if you simply state that the limit of 0.999... as we keep adding more decimal places is 1. I think this is also a closer translation of the mathematical statement into plain english.

I think that is exactly my point. If we call limits being equal to, in the equation for derivatives, lim Δx→0 (f(x+Δx)-f(x))/Δx (I know I may not have written it right), the answer would be undefined.

 

[WWGD]

They are clearly different but if you're approximating, they're almost equal.

No, they are not different as Real numbers using the definition of Real numbers in terms of Cauchy sequences. This is a technical point that requires the use of precise definitions.

 

[mathwonk]

this question never dies. this is basically why people cannot tell the difference between a good car and a car that says "BMW" on the side, i.e. its not all about the notation. but for some reason they are willing to allow the same car to be called a bmw or a beemer, but a number can't have two names. still the same people who admit that .33333... = 1/3, will not accept the result of multiplying this equation by 3.

 

[WWGD]

I agree with Mathwonk. It is a very narrow view of what a number is. A number is not a unique object, but instead an equivalence class of objects. Under this equivalence relation, the two, 0.99999... and 1 are equivalent, and so considered equal. Maybe similar to the fact that if today is Wednesday, then every day that is $\pm 7k$ days from today is also a Wednesday, and not just that today , 12/9/2015 is Wednesday, so no other day can be a Wednesday . Equivalence is very close conceptually to equality but it is not quite the exact same thing.

 

[axmls]

I think a key point is that .999... is by definition a limit, and that limit is equal to 1.

 

[WWGD]

I think a key point is that .999... is by definition a limit, and that limit is equal to 1.

True, it is the limit of a Cauchy sequence, which converges by the fact that the Reals are a complete metric space, and the sequence, as you said, converges to 1.

 

[Mark44]

I would think this whole theme is answered by:
The definition of real numbers says that $$0.999..._{\text{as real number}} \ =_{\text{acting on real numbers}} \ 1_{\text{as real number}}$$
If you don't like it, use other numbers.

I have no idea what you're trying to say in the equation above.

I think equivalence is more intuitive if you simply state that the limit of 0.999... as we keep adding more decimal places is 1. I think this is also a closer translation of the mathematical statement into plain english.

There's a difference between what you say above and what I think you're trying to say.
First: limit of .999 ... is meaningless, as you aren't saying what is changing.
Second, $\lim_{n \to \infty}\sum_{j = 1}^n \frac 9 {10^j} = 1$. The sum can be made arbitrarily close to 1 just by taking more terms in the sum.

 

[Mark44]

I have no way to say this is true for sure, but it appears to me as though proof #3 is based off an incorrect assumption. Wouldn't the difference between .999... and 1 be infinitesimal, not 0.

There is no need that I can see to bring infinitesimals into the argument. By definition of the notation, .999 ... means an infinitely long string of 9s. If there is a difference between .999 ... and 1, there has to be a digit somewhere in the difference that isn't 0. Which digit is that?

Isn't this one of the reasons we have the concept of infinity, and in this case an infinitesimal value. If something infinitesimal is just considered 0, wouldn't a Riemann sum not mater at all, and just be equal to zero? Wouldn't that make anything using dx, dy, etc. represent a change of zero, as they are defined as an infinitesimaly small change in their respective variable? This is just my thinking, I am not being ignorent, I just want to understand.

 

[WWGD]

I have no way to say this is true for sure, but it appears to me as though proof #3 is based off an incorrect assumption. Wouldn't the difference between .999... and 1 be infinitesimal, not 0. Isn't this one of the reasons we have the concept of infinity, and in this case an infinitesimal value. If something infinitesimal is just considered 0, wouldn't a Riemann sum not mater at all, and just be equal to zero? Wouldn't that make anything using dx, dy, etc. represent a change of zero, as they are defined as an infinitesimaly small change in their respective variable? This is just my thinking, I am not being ignorent, I just want to understand.

Like someone pointed out, there are no infinitesimals in the Standard Real numbers; these exist only on non-standard models of the Reals. Your other questions are good, let me think about them some more. I think the point is that the rectangles, while going to 0, are uniformly bounded , by compactness of the region of integration, so these widths do not become indefinitely-close to 0; they become close to 0 in a controlled way. Good point, though.

 

[Mark44]

If something infinitesimal is just considered 0, wouldn't a Riemann sum not mater at all, and just be equal to zero? Wouldn't that make anything using dx, dy, etc. represent a change of zero, as they are defined as an infinitesimaly small change in their respective variable? This is just my thinking, I am not being ignorent, I just want to understand.

In the Riemann definition of a definite integral, you have the number of rectangles n approaching infinity (i.e., increasing without bound), where the area of each rectangle is approaching zero.
In effect, what you have is the indeterminate form $[0 \cdot \infty]$. Indeterminate forms are a way to categorize limits that involve a product or quotient or other operation in which one or more of the quantities is unbounded. Indeterminate forms are not number, and different instances of the same indeterminate form can come out to be literally any real number or $\infty$ or $-\infty$. With Reimann sums, if you have a well-behaved function, the limit comes out to a real number.

Like someone pointed out, there are no infinitesimals in the Standard Real numbers; these exist only on non-standard models of the Reals. Your other questions are good, let me think about them some more. I think the point is that the rectangles, while going to 0, are uniformly bounded , by compactness of the region of integration, so these widths do not become indefinitely-close to 0; they become close to 0 in a controlled way. Good point, though.

See above...

 

[WWGD]

In the Riemann definition of a definite integral, you have the number of rectangles n approaching infinity (i.e., increasing without bound), where the area of each rectangle is approaching zero.
In effect, what you have is the indeterminate form $[0 \cdot \infty]$. Indeterminate forms are a way to categorize limits that involve a product or quotient or other operation in which one or more of the quantities is unbounded. Indeterminate forms are not number, and different instances of the same indeterminate form can come out to be literally any real number or $\infty$ or $-\infty$. With Reimann sums, if you have a well-behaved function, the limit comes out to a real number.
See above...

Yes, what I meant is for a specific Riemann sum, the widths are uniformly bounded; the function Min width is a continuous function on a compact set -- integration is done on a compact interval [a,b] -- so it assumes an actual minimum value on [a,b]. Now, we can consider what happens as we let the widths go to 0, but for a _specific_ Riemann sum, the widths do not go to 0. And then we define convergence using nets.

 

[Mark44]

.

Yes, what I meant is for a specific Riemann sum, the widths are uniformly bounded; the function Min width is a continuous function on a compact set -- integration is done on a compact interval [a,b] -- so it assumes an actual minimum value on [a,b]. Now, we can consider what happens as we let the widths go to 0, but for a _specific_ Riemann sum, the widths do not go to 0.

Sure they do. Assuming for simplicity's sake that $\Delta x = \frac{b - a} n$, as $n \to \infty, \Delta x \to 0$. There is no positive minimum value that's a lower bound on $\Delta x.$

For the same reason, we can evaluate $\lim_{x \to \infty} x \cdot \frac 1 {2x + 3}$ and get 1/2, even though there is no upper bound on x, and no positive lower bound on $\frac 1 {2x + 3}$.

And then we define convergence using nets.

 

[WWGD]

.Sure they do. Assuming for simplicity's sake that $\Delta x = \frac{b - a} n$, as $n \to \infty, \Delta x \to 0$. There is no positive minimum value that's a lower bound on $\Delta x.$

For the same reason, we can evaluate $\lim_{x \to \infty} x \cdot \frac 1 {2x + 3}$ and get 1/2, even though there is no upper bound on x, and no positive lower bound on $\frac 1 {2x + 3}$.

Yes, but a specific Riemann sum takes on a specific value for $n$. Then you consider different Riemann sums where n becomes larger. So you have a collection of Riemann sums, each with a specific value of n, and we consider what happens as we consider specific sums with increasingly-larger values for $n$. So you end up with a net of Riemann sums , where the order is inclusion of partition set {#x_1,x_2,...,x_k#} , that converges ( under the right conditions) as a net to the value of the integral.

 

[Mark44]

Yes, but a specific Riemann sum takes on a specific value for $n$. Then you consider different Riemann sums where n becomes larger.

But the Riemann integral is the limit as the partition gets finer, of these sums.
https://en.wikipedia.org/wiki/Riemann_integral

Loosely speaking, the Riemann integral is the limit of the Riemann sums of a function as the partitions get finer.

So you have a collection of Riemann sums, each with a specific value of n, and we consider what happens as we consider specific sums with increasingly-larger values for $n$.

In other words, as you take the limit as n goes to infinity, which forces the subinterval width to zero.

 

[WWGD]

But the Riemann integral is the limit as the partition gets finer, of these sums.
https://en.wikipedia.org/wiki/Riemann_integral

In other words, as you take the limit as n goes to infinity, which forces the subinterval width to zero.

This is kind of difficult to explain in words, but my point is that, for any individual Riemann sum, the function (partition width) is a continuous function on a compact set [a,b], so it assumes a minimum value. Of course, we let the minimal values go to 0.

 

[Mark44]

This is kind of difficult to explain in words, but my point is that, for any individual Riemann sum, the function (partition width) is a continuous function on a compact set [a,b], so it assumes a minimum value. Of course, we let the minimal values go to 0.

A particular Riemann sum will involve a finite number of subintervals of nonzero width. The Riemann integral, however, is the limit of these Riemann sums, as the subinterval width goes to zero.

 

[WWGD]

A particular Riemann sum will involve a finite number of subintervals of nonzero width. The Riemann integral, however, is the limit of these Riemann sums, as the subinterval width goes to zero.

So we are saying the same thing. I was addressing a poster question about the fact that an indefinitely-small Standard Real number must equal 0, which led them to believe that it must follow that a Riemann sum must equal 0 , since all the terms in the sum become indefinitely small. But this is not so, because in an individual/specific Riemann sum, the terms do not become indefinitely small; they are uniformly bounded by my previous argument. Of course this bound goes to 0, but the minimal width on a specific Riemann sum does not go to 0 or else the sum itself would be zero. Now, it would be interesting how to do integrals in non-standard analysis. I have seen methods to differentiate but I have not seen integration theory in non-standard analysis.

 

[Mark44]

I'm not sure that we were. Isaac0427 said

wouldn't a Riemann sum not mater at all, and just be equal to zero?

I believe he really meant Riemann integral here.

Your response to him was

I think the point is that the rectangles, while going to 0, are uniformly bounded , by compactness of the region of integration, so these widths do not become indefinitely-close to 0; they become close to 0 in a controlled way.

Your explanation used a lot of fairly high-powered analysis, but was flawed; i.e., that the (areas of) the rectangles are going to zero while their widths miraculously remain positive, or that they "become close to zero in a controlled way."

 

[WWGD]

I'm not sure that we were. Isaac0427 said I believe he really meant Riemann integral here.

Your response to him was
Your explanation used a lot of fairly high-powered analysis, but was flawed; i.e., that the (areas of) the rectangles are going to zero while their widths miraculously remain positive, or that they "become close to zero in a controlled way."

Do you see something specifically wrong with the claim that rectangle width is a continuous function on a compact set so it achieves a minimum, which is a uniform lower bound? Is this _not_ a controlled way? What flaw do you see here? Or do you claim that in a specific Riemann sum the widths, and specifically the minimal width, becomes indefinitely close to 0. Convince me then that the integral is not zero.

 

[Mark44]

Do you see something specifically wrong with the claim that rectangle width is continuous on a compact set so it achieves a minimum, which is a uniform lower bound? Is this _not_ a controlled way? What flaw do you see here?

This part, which I just quoted:

so these widths do not become indefinitely-close to 0;

 

[WWGD]

This part, which I just quoted:

In a _specific sum_ , the function interval width is continuous and defined on a compact set. What is wrong with this argument? If you let your minimal width go to 0 in a _specific sum_ without bound, your integral will equal 0. What you end up getting is a net of Riemann sums, which will converge netwise to the value of the integral under the right conditions.

 

[WWGD]

I also see it like this: The collection {$x_1=a, x_2,..., x_n=b$} covers [a,b]. By compactness of [a,b], the collection has a finite subcover. Now, consider min $(x_{i+1}-x_i )$:=m on the finite subcover. A finite set will have a minimum value , say m. The value $m=0$ will never be realized. Maybe I am not explaining myself well here.

 

[WWGD]

Sorry, could not edit. Just to say I need to get to sleep, I will address any other disagreement tomorrow.

 

[Mark44]

In a _specific sum_ , the function interval width is continuous and defined on a compact set. What is wrong with this argument? If you let your minimal width go to 0 in a _specific sum_ without bound, your integral will equal 0. What you end up getting is a net of Riemann sums, which will converge netwise to the value of the integral under the right conditions.

We are not talking about a specific Riemann sum. My complaint is that you are talking about a limit, but don't seem to realize it. Emphasis added.

I think the point is that the rectangles, while going to 0, are uniformly bounded , by compactness of the region of integration, so these widths do not become indefinitely-close to 0

In other words, as $\Delta x \to 0$ -- that's a limit. All the other stuff about compact sets, continuity, and subcovers etc. is just word salad that I'm 99% sure would be over the head of the person you were responding to.

 

[madness]

There's a difference between what you say above and what I think you're trying to say.
First: limit of .999 ... is meaningless, as you aren't saying what is changing.
Second, $\lim_{n \to \infty}\sum_{j = 1}^n \frac 9 {10^j} = 1$. The sum can be made arbitrarily close to 1 just by taking more terms in the sum.

But what I actually said was "limit of 0.999... as you keep on adding more decimal places", which is not the same as what you quoted from me. I'm not sure what point you are making in the second part, as it seems to agree with what I wrote. The mathematical statement simply says that as you make n bigger and bigger you get closer and closer to 1, which I don't think anybody would disagree with.

 

[PeroK]

But what I actually said was "limit of 0.999... as you keep on adding more decimal places", which is not the same as what you quoted from me. I'm not sure what point you are making in the second part, as it seems to agree with what I wrote. The mathematical statement simply says that as you make n bigger and bigger you get closer and closer to 1, which I don't think anybody would disagree with.

It is important, though, to realize that a limit is a number, and not a process. It's a small point, perhaps, but it sharpens your understanding of these things.