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According to Rosen, an infinite set A is countable if $|A|= |\mathbb{Z}^+|$ which in turn can be established by finding a bijection from A to $\mathbb{Z}^+$.

Also, a sequence is defined as a function from $\mathbb{Z}^+$ (or $\{0\} \cup \mathbb{Z}^+$) to some set.

With the above, a sequence is certainly enumerable. However, it need not be a bijection, e.g. Fibonacci(1) = Fibonacci(2) = 1.

This implies that not every sequence is countable which seems counterintuitive. Are there any results in this regard? Is there a mistake in the reasoning above?

wsaleem
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  • Is the sequence $1,1,1,...$ countable? – Yanko Feb 07 '18 at 17:30
  • I suppose you mean the values taken by the sequence, that is, given a sequence $(a_n)_{n \in \mathbb N}$ (or $\mathbb Z^+$) then you're asking if the set ${a_n : n \in \mathbb N}$ is countable, right? – amrsa Feb 07 '18 at 17:34
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    "This implies that not every sequence is countable" ... um, i'm not following. Why does it imply that? A function from a countable set must be countable whether it is a bijection or not. – fleablood Feb 07 '18 at 17:35
  • @amrsa Yes, that is the question. – wsaleem Feb 07 '18 at 18:05
  • @Yanko The set of terms is {1} which is finite, thus countable. – wsaleem Feb 20 '18 at 21:59
  • In that case every sequence is countable. – Yanko Feb 21 '18 at 13:35

3 Answers3

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Every sequence has a countable or a finite set of values.

Besides, you are mixing two ideas : a sequence $(u_n)_n$ is a function $n\mapsto u_n\in F$ ($F$ being any possible set) and almost never a bijection, but the set of all its values are finite or countable.

Netchaiev
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"However, it need not be a bijection"

No, it doesn't.

"This implies that not every sequence is countable"

Why do you say that?

$f: \mathbb Z^+ \to B$. If $f$ is not surjective then there are points of $B$ that are not in the image. Those to not matter. We can restrict ourselves to $f: \mathbb Z^+ \to Im(B)$.

This must be surjective.

It doesn't need to be injective however and your Fibinocci example shows.

But... so what? Than means $|Im(B)| \le |\mathbb Z^+|$.

Hence it MUST be countable (or countably finite).

Anyway, as the terms of a sequence, as opposed to a set, need not be distinct it is possible, indeed common, for a sequence to have a finite number of distinct terms infinitely repeated.

fleablood
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  • A bijection is invertible. Consider the bijection from $\mathbb{Z}^+$ to the terms of the Fibonacci sequence. If the set of terms of the Fibonacci sequence is countably infinite, then, as per definition, such a bijection must exist. Let us call it $f$. What is $f^{-1}(1)$? – wsaleem Feb 07 '18 at 18:10
  • @wsaleem The bijection need not be given by the sequence itself ... – Noah Schweber Feb 07 '18 at 18:13
  • Agreed. But no matter the function, we will have 2 distinct elements in the domain ($\mathbb{Z}^+$) mapping to the element, 1, in the co-domain. This function is not injective and therefore not a bijection. Thus, it seems that no bijection exists from $\mathbb{Z}^+$ to the set of terms of the Fibonacci sequence. – wsaleem Feb 07 '18 at 18:17
  • Wsaleem. I honestly fail to see your point. – fleablood Feb 08 '18 at 00:25
  • f (1)=13,f(2)=8;f (3)=1;f (4)=21;f (5)=3 etc. Is a bijection. – fleablood Feb 08 '18 at 00:31
  • A sequence may list a value many times that the range set itself is an unordered set of distinct values. Take $c_n = n^2* \frac 16 - \lfloor n^2* \frac 16 \rfloor ={\frac 16, \frac 23, \frac 12, \frac 23, \frac 16, 0, \frac 16, etc}$. This is a map from $\mathbb N\to{0,\frac 16,\frac 23,\frac 12}$ a finite set. This is clearly not a bijection. $c^{-1}(\frac 23) = {2,4,8,10,14,16, etc}$. The set ${0,\frac 16,\frac 23,\frac 12}$ is not the same thing as the sequence which lists the set in some order with repetition. – fleablood Feb 08 '18 at 01:04
  • The set of Fibonacci numbers is ${1,2,3,5,8....}$. $1$ is not in the set twice. – fleablood Feb 08 '18 at 01:04
  • The "set of terms of the Fibonacci sequence is the set ${1,2,3,5,8,13,etc}$ there are many many bijections to it. But the one function we use to list them $f_1 = 1; f_2 = 1; f_n = f_{n-1} + f_{n-2}$ does not happen to be a bijection. So what? There are others that are. Example $g(1) =1; g(2) =2; g(3)=3; g(4)=5.... g(n) = f_{n+1}$ is a bijection. – fleablood Feb 08 '18 at 01:11
  • "But no matter the function, we will have 2 distinct elements in the domain (Z+) mapping to the element, 1, in the co-domain." Not at all true. Why do you think that? Just because $1$ is listed twice in the sequence does not mean it must be mapped to twice in a function. The set of fibonnaci is not the same thing as the sequence and just because a sequence may list terms multiple times, the only exist in the set once. – fleablood Feb 08 '18 at 01:13
  • Sorry, @fleablood, I did not receive notifications about your comments so was unaware. I think I see your point. – wsaleem Feb 20 '18 at 22:00
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As a sequence is a set indexed by the natural numbers, there exists a surjection from the naturals to the set. Let $A$ be the set and g: $\mathbb{N}\rightarrow A$ the surjection. Then we can define a map $f:A \rightarrow \mathbb{N}$ as $f(a)=$ the minimum natural number n such that $g(n)=a$. Then f is injective and thus A must be countably infinite or finite. Further, you can similarly show that every set indexed by a countably infinite set is either finite or countably infinite.