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January 28th, 2016, 12:22 PM   #21
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Originally Posted by Pengkuan View Post
In this case, the list of reals, for example your $\pi a^{-n}$, will create a real number, which we append to the list. Then there is no more diagonal.
No, we don't append to a list, we create an infinite list of reals of the form $\pi a^{-n}$. There are an infinite number of such reals, one per natural number.

Having said that we can add more real numbers to the list if we like. There's always a diagonal.
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January 28th, 2016, 12:33 PM   #22
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No, we don't append to a list, we create an infinite list of reals of the form $\pi a^{-n}$. There are an infinite number of such reals, one per natural number.

Having said that we can add more real numbers to the list if we like. There's always a diagonal.
Do you mean that once an infinite list is created, no new member can be added?
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January 28th, 2016, 12:55 PM   #23
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No. Bit it doesn't make the list any longer. It's still an infinite list.
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January 28th, 2016, 01:04 PM   #24
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No. Bit it doesn't make the list any longer. It's still an infinite list.
You said "we don't append to a list" and "Bit it doesn't make the list any longer"

First you say We do not change the list, then you say even it is changed, it is not any longer.

If the list cannot be changed, it leads to a thing, if it change, it leads to another thing. What is your mind. It changes or not?
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January 28th, 2016, 06:55 PM   #25
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The diagonal proof is not about building an infinite list, it just says that we should assume that we have one.

But if you want to add entries to an infinite list, you can. But it doesn't make the list any longer.
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January 28th, 2016, 07:27 PM   #26
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Originally Posted by Pengkuan View Post
leads to a thing, if it change, it leads to another thing. What is your mind. It changes or not?

It changes but it still has the same cardinality.

Suppose I have an infinite set X = 1, 2, 3, 4, ...

Now suppose I "add" an element to the end to get Y = 1, 2, 3, 4, ..., 0.

But this second set has the same cardinality as the first one, and the bijection is to just map 0 to 1, 1 to 2, 2 to 3, etc.

Y - X
0 - 1
1 - 2
2 - 3
etc.

So you see we've added a new element to X to get Y, yet we have not changed the size.
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January 29th, 2016, 06:48 AM   #27
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The diagonal proof is not about building an infinite list, it just says that we should assume that we have one.

But if you want to add entries to an infinite list, you can. But it doesn't make the list any longer.
So, the length of the list is infinity. When added one, it is still infinity. The number of digits is none-changed infinity. And is the diagonal unchanged, or not? THe added sequence is in the diagonal or not?
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January 29th, 2016, 06:49 AM   #28
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It changes but it still has the same cardinality.

Suppose I have an infinite set X = 1, 2, 3, 4, ...

Now suppose I "add" an element to the end to get Y = 1, 2, 3, 4, ..., 0.

But this second set has the same cardinality as the first one, and the bijection is to just map 0 to 1, 1 to 2, 2 to 3, etc.

Y - X
0 - 1
1 - 2
2 - 3
etc.

So you see we've added a new element to X to get Y, yet we have not changed the size.
It is about the diagonal. If we add one sequence and still have the same length, the diagonal stays the same.
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January 29th, 2016, 07:21 AM   #29
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It is about the diagonal. If we add one sequence and still have the same length, the diagonal stays the same.
How can it be the same? It has a digit from the new row that we added.
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Originally Posted by Pengkuan View Post
So, the length of the list is infinity. When added one, it is still infinity. The number of digits is none-changed infinity. And is the diagonal unchanged, or not? THe added sequence is in the diagonal or not?
Do not think of "infinity" as a number, it's not. "Infinite" is a property that means "non-terminating". So, if we add an entry to a non-terminating list, it is still non-terminating. It also has another entry that it didn't have before.

But we don't run out of columns with which to build the diagonal, because the horizontal list of columns is also non-terminating.

Last edited by v8archie; January 29th, 2016 at 07:27 AM.
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