What is the explanation behind the diagonal (pseudo)paradox in mathematics?

[radou]

http://mathworld.wolfram.com/DiagonalParadox.html

the approximation of the diagonal with a finite nuber of line steps only appears (to the eye) as the real diagonal for a very large n, but as long as we consider it an approximation, it's length (the summation of horizontal and vertical steps) is always be 2. it's only a problem of your point of view, so i don't really see a paradox here..does someone have a better idea?

 

[matt grime]

In the limit as the number of steps tends to infinity the "step" converges to the diagonal. Of course "converge" requires a distance notion, but in any reasonable notion of the distance between to functions they do indeed converge. The "paradox" is just that the measurements of length of the two lines do not agree, ie you cannot interchange limits and lengths. as with all paradox it only contradicts what "most" people expect, ie it is surprising that we can't interchange them.

 

[radou]

yup, that was my point :) so, i guess the wolfram crew should mention these few facts :)

 

[matt grime]

Another way of stating what is going on generally.

Suppose that X(n) is a sequence of objects that have a meaningful limit X. If all of hte X(n) have a property P, then when does X have property P. There are numerous theorems in maths that follow this pattern.

the paradox here is that most people will accept that the limit must have P without thinking about. there are indeed many people who use such fallacious reasoning to claim things in maths that are patently absurd.

 

[saltydog]

In the limit as the number of steps tends to infinity the "step" converges to the diagonal.

Hello Matt. I do not relish the though of taking issue with you but feel the step never converges to the diagonal. This is my argument: The total length of step is:

n(2)(\frac{1}{n})

So even as n goes to infinity, the length of steps remains at 2.

However, the sum of all the diagonals of these steps is:

a(\frac{\sqrt{2}}{a})

And that expression is always \sqrt{2} even as a tends to infinity.

Would you kindly provide a proof showing that the step function as defined in the Wolfram document converges to the diagonal.

 

[master_coda]

Hello Matt. I do not relish the though of taking issue with you but feel the step never converges to the diagonal. This is my argument: ...

Didn't Matt just make two postings pointing out that it's wrong to assume that a sequence of curves of length 2 must necessarily converge to a curve of length 2?

 

[matt grime]

When i said "of course converge requires a distance notion, but in any reasonable notion of the distance between to functions they do indeed converge" i meant it. in what sense do you think they don't converge, saltydog? pointwise? the sup norm? the "l^1" norm (ie the integral of the area between them)? (the step functioon isn't admittedly a a cont function from R to R at first glance, but twist the picture 45 degrees so it's a sawtooth if you need to)


some properties do pass through limits some do not, it is important to fidn out which ones can be presumed to do so.

 

[saltydog]

Hello Matt. Glad you didn't exhibit offense at my insolence. I am in error sir and yield to you.

My disagreement stemed from my view that the arc lengths of both the diagonal and the step function never seem to match. However your statement that the area between the two necessarilly goes to zero is intutitive justification for me that they converge. However, this seems to be point-wise convergence.

However, I still have a problem resolving the conflict with the two arclengths: It seems to me that even though the step function converges to the diagonal, the arc-length always remains at 2 whereas the diagonal is root(2). Is this due to the nature of point-wise convergence or to my lack of familiarity with the subject?