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I assume that $\frac{1}{3}$ is equal to $0.3333...$.

Let's define a sequence as follows: $0.3$, $0.33$, $0.333$, $0.3333$,...

Question: is $\frac{1}{3}$ included in this sequence?

Every item in the sequence clearly has finite number of decimals, and $\frac{1}{3}$ has infinite decimals so it is clearly not included.

On the other hand the limit of this sequence is $\frac{1}{3}$ so it seems ok to say that it "includes" $\frac{1}{3}$.

What is the correct answer?

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    $\begingroup$ The correct answer is that it is not included. Every entry in the sequence is smaller than the limit. $\endgroup$
    – Peter
    Apr 13 at 9:39
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    $\begingroup$ is $\frac{1}{3}$ included in this series? You need to defined what "included" here means. As a limit, yes. $\endgroup$ Apr 13 at 9:39
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    $\begingroup$ What you mentioned is not a series , but a sequence. A series is a summation of entries of a sequence. And a better formulation would be "belongs to" instead of "included". $\endgroup$
    – Peter
    Apr 13 at 9:45
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    $\begingroup$ @ForwardEd Physicists are well aware that, in the real world, "very close" is generally indistinguishable from "equal". $\endgroup$
    – David Z
    Apr 14 at 9:29
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    $\begingroup$ @user3840170: don't be ridiculous. We all know what the OP means. $\endgroup$
    – TonyK
    Apr 14 at 23:36

4 Answers 4

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One thing that hasn't been said in the previous answers, and really needs to be emphasized, is that mathematics is not about guesswork, so we really cannot think "so it seems ok to ...". In mathematics, everything has to be precisely defined, and then whatever (precise) statement you make would be either true or false, and this is not up for debate.

Every item in the sequence clearly has finite number of decimals, and $1/3$ has infinite decimals so it is clearly not included.

Although not totally precise, you are essentially correct here. Precisely, each item in the sequence is equal to $k/10^m$ for some natural numbers $k,m$, but $1/3$ is not. We can prove this: Given any $k,m∈ℕ$ such that $1/3 = k/10^m$, we have $10^m = 3·k$, but $10^m = (9+1)^m$ $≡ 1^m = 1$ (mod $3$), whereas $3·k ≡ 0$ (mod $3$), so this is impossible.

On the other hand the limit of this sequence is $1/3$ so it seems ok to say that it "includes" $1/3$.

You can very well define "include" to mean something in order to make that sequence include its limit. But you must be very clear about the fact that it is not the conventional meaning of "include". So you were correct to use scare-quotes around "include", and it is important to realize how crucial precision is in mathematics.

Here is a standard mathematical notion that you may be interested in. Given any set $S ⊆ ℝ$, we say that $c$ is an adherent point of $S$ iff $c$ is a limit of some sequence from $S$. Given any sequence $f∈ℕ→ℝ$ and any $c∈ℝ$, it turns out that the following are equivalent:

  • $c$ is an adherent point of $\{ \ f(k) : k∈ℕ \ \}$.
  • There is some sequence $i∈ℕ→ℕ$ such that $f∘i$ converges to $c$.
  • Either there is some increasing sequence $i∈ℕ→ℕ$ such that $f∘i$ converges to $c$ or there is some $k∈ℕ$ such that $f(k) = c$.

This notion may be the kind of relationship you are interested in, but it is not a good idea to call it "inclusion".

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    $\begingroup$ Although it makes sense to say that in this context, but mathematics is about guesswork, and a large part of mathematics starts out from guesswork. Proofs and definitions are a necessary evil to make sure we communicate with each other and don't go crazy. They are not inherently a part of maths. $\endgroup$
    – Trebor
    Apr 15 at 4:01
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    $\begingroup$ @Trebor: Ok ok.. let's compromise. Rigorous mathematics is not about guesswork, though one is free to guess as much as they like about what statements to try proving and what proof ideas to try. $\endgroup$
    – user21820
    Apr 15 at 10:13
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That is a series which converges to $\frac{1}{3}$ which, informally, means it will eventually get arbitrarily close. However, no entry in the series will be exactly $\frac{1}{3}$.

Similarly, the series $3$, $3.1$, $3.14$, $3,141$, $3.1415$ $3.14159$, ... converges to $\pi$ but no term is equal to $\pi$.

As Peter says in the comments, it would be better to say "sequence".

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A single term of your series can be expressed as

$$A_n=\frac{1}{3}\bigg(1-\frac{1}{10^n}\bigg)$$

Using basic limits, this means that if $n→∞,A_n→\frac{1}{3}$. But according to your definition, $n$ is a countable number. Hence logically,$A_n<\frac{1}{3}$.

If you consider including the value even when n is $\infty$, then you may say it is included in the series, which isn't possible though. Like it depends upon your definition of "inclusion". Hope I was able to answer your question

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  • $\begingroup$ So if the sequence has countably infinite elements then it would be correct to say that that sequence lists 1/3 because 1/3 also has countably infinite decimals? $\endgroup$
    – psmith
    Apr 13 at 9:56
  • $\begingroup$ What do you mean by "countably infinite" @psmith $\endgroup$
    – Gwen
    Apr 13 at 9:57
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    $\begingroup$ A set is countably infinite if its elements can be put in one-to-one correspondence with the set of natural numbers. Each term in your series you provided can be corresponded to each n being a natural number. But you can't say that "∞" is included in the set of all natural numbers, can you? Similarly you actually can't say that "1/3" is a member of your series. Check my edited answer. $\endgroup$
    – Gwen
    Apr 13 at 10:08
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    $\begingroup$ @psmith No, definitely not. Your sequence has infinitely many elements of the form $0.333\ldots3$, with finitely many $3$s, but it contains no element which is $0.333\ldots$, with infinitely many $3$s. $\endgroup$ Apr 14 at 12:54
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    $\begingroup$ @psmith Why would we be able to conclude that? The existence of infinitely many finite things does not entail the existence of a single infinite thing. $\endgroup$ Apr 16 at 10:38
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As others have pointed out, this is a matter of definition. We might translate the statement that 1/3 is included in the sequence to something a little more precise, such as:

There exists an index such that the element of the sequence at that index is equal to 1/3.

In this case, if we're indexing by natural numbers, then the answer is NO (because it can be shown that every ℕ-indexed element of the sequence is distinct from 1/3).

However, a different choice would be to permit the index ω of a convergent sequence, in which case, the answer is YES, because it can be shown that the sequence converges to 1/3, and, by our definition, the element of the sequence at ω is that limit.

Now, this latter view is, I think, unconventional, but it is a choice that could be made.

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    $\begingroup$ Um. And what are the elements around that one? If you allow ω, clearly you must allow ω+1 and ω-1. $\endgroup$
    – Sneftel
    Apr 15 at 10:53
  • $\begingroup$ @Sneftel i guess he is talking about the ordinal number ω (omega) that is greater than every natural number. $\endgroup$
    – psmith
    Apr 15 at 15:13
  • $\begingroup$ @psmith Exactly. The "infinity" in ordinals isn't the same as the infinity in the limit of a sequence. If you never get there, you never get there. $\endgroup$
    – Sneftel
    Apr 15 at 15:55
  • $\begingroup$ @Sneftel Clearly? Again, it's a matter of definition. In this case, I have chosen to allow precisely a natural number, or a symbol ω as an index; nothing more. I do not require this index set to be a Rig, only to be totally ordered. Like I said, I don't claim this is a common definition, but it is a choice that could be made. $\endgroup$ Apr 16 at 14:06
  • $\begingroup$ Sure. For that matter, you could simply make the choice that in this new system, the seventh element of the sequence is exactly 1/3, arithmetic be damned. My point is that somebody who would actually do math using this would not be able to do useful math if they had made either of those choices, because of the loss of consistency in the axioms underlying the math. $\endgroup$
    – Sneftel
    Apr 16 at 14:19

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