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# What counts as a legitimate sequence?

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Hello,

I'm in a online debate with someone about some deep mathematical concepts. My opponent was trying to convince me that you can have a sequence of numbers for which there is a first element, a last element, but no second element and no second last element (where the sequence contains more than 2 elements). I thought that was absurd until he gave me an example: all the real numbers between 0 and 1.

It definitely has a first member (0) and it definitely has a last member (1), but after 0 there is no "next" real number. Likewise, there is no real number that comes just before 1. Yet there are obviously real numbers between 0 and 1.

That stumped me until I figured that couldn't possible count as a sequence because sequences must consist of well-define discrete elements and real numbers aren't well-defined or discrete. I thought that was a terrible way of putting it, so I looked up the definition of sequences online and the key word I found was "enumerable". The members of a sequence must be enumerable. And I don't believe the reals are enumerable.

But then he came up with this other example: take the sum $$\sum_{i=1}^{n}\frac{9}{10^i}$$. If you define each member of the sequence as the value of this sum for every value for n > 0 and order them by each incremental value of n, then you will have the sequence (0.9, 0.99, 0.999, ...). And if you allow n = $$\infty$$, then we know this sum equals 1. Therefore, 1 is the last value in the sequence. Therefore, the sequence starts with 0.9 and ends with 1. Furthermore, each member is well-defined and discrete. We know each member by the sum $$\sum_{n}{i^1}\frac{9}{10^i}$$ and the value of n. Yet, it has no second last member.

Is this a legitimate example of a sequence?

Edited by gib65
trying to get latex right

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19 minutes ago, gib65 said:

Yet, it has no second last member.

Surely this is true of any infinite sequence?

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Just now, Strange said:

Surely this is true of any infinite sequence?

Well, the point is, it has a last without a second last.

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23 minutes ago, gib65 said:

Therefore, 1 is the last value in the sequence.

No, it is in an infinite sequence. There is no last value. Values approach 1, in the limit, but never get there.

You can calculate the Nth digit of Pi (without having to calculate all the previous digits) but you can't calculate the second to last. Because there is no last.

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A sequence of real numbers is a mapping of the positive integers (or, in the case of a finite sequence, 1 to some n) to the real numbers.  That is, "1" is assigned some number so that number is the first number in the sequence.  "2" is assigned to some number so that is the second number in the sequence.

The numbers in the interval [0, 1] is not a sequence.

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So would you say that the sequence (0.9, 0.99, 0.999, ... , 1) is not a sequence since 1 can't be mapped?

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3 hours ago, gib65 said:

And if you allow n = , then we know this sum equals 1

Why should you 'allow' it ?

This sort of thing is exactly why n is not equal to infinity, whatever that means.

Some writers are more particular and strict about specifying sequences such as 0., 0.99, 0.999, ... ( are you aware to the 3 dots convention?) than others are.

n is always finite, but never terminates. ( ie there is no greatest n)

And it is this no greatest property that we use here.

I suggest you look up the definition of a 'Cauchy sequence'

See also this thread here

Edited by studiot

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The concept of sequence is well defined - a set which is 1-1 with integers (all or part).  What is being described are sets which are not necessarily sequences.

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22 hours ago, gib65 said:

And if you allow $$n = \infty$$, then...

As studiot points out, you don't. Or rather, you shouldn't.

The symbol $$\infty$$ is used in various roles in math. Most often as part of a standard notation, such as in $$\sum_{i=1}^\infty$$ which is shorthand for $$\lim_{n\to \infty} \sum_{i=1}^n,$$ which is itself a shorthand for a more complicated expression that defines the value of such a limit (and which happens to no longer contain the $$\infty$$ symbol). In this way the use of $$\infty$$ in the expression $$\sum_{i=1}^\infty$$ is not much different from the use of the Greek letter $$\Sigma$$ in the same expression. It doesn't stand for' anything, in the same way that $$n$$ does in the similar looking $$\sum_{i=1}^n,$$ where $$n$$ represents a natural number value.

There are several contexts when $$\infty$$ is the symbol chosen to represent some fixed object of particular importance. Most famously when extending the reals $$\mathbb{R}$$ by declaring that $$\infty$$ be some object that is not itself a real, and that the usual ordering $$<$$ of $$\mathbb{R}$$ get extended so that $$\infty$$ is larger than each real number in a new larger ordering.

At least a dozen other contexts have the symbol $$\infty$$ representing objects of special properties. It means that next time you board a plane and sit down in your assigned seat next to a person completely unknown to you, and you think that you should initiate conversation by making some observations on the properties of $$\infty,$$ you do have to be specific about the context. If the other fellow happens to be specialist in the theory of elliptic curves, they may not realize you are talking about the $$\infty$$ which is used to compactify the complex plane, and they might think, judging from your remarks, that you are a bit weird.

Everyday language has the same ambiguities, where the same word may have different meanings. Cf. the quote from Mae West: "Women like a man with a past, but they prefer a man with a present." And other examples exist. I am not aware of any language crackpots' who get irate about this fact and are willing to denounce the use of language as defective and useless. In contrast, math crackpots have a field day with examples such as $$\infty$$ that can take different meanings depending on context.

The point is that if your context of choice already assumes that $$\infty$$ is the name of a special object, then for all we know it might be that $$\infty$$ is fixed at the natural number value $$42.$$ In that case you seem justified to let $$n = \infty,$$ in which case follows $$\sum_{i=1}^\infty a_i = \sum_{i=1}^{42} a_i.$$ This is bad, because then you are using the expression $$\sum_{i=1}^\infty$$ in two incompatible ways, which makes it `ill-defined', we would not know what it means, and therefore it means nothing.

Edited by taeto

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