# Is the set of terms of a sequence countable?

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?

• Is the sequence $1,1,1,...$ countable? Feb 7, 2018 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? Feb 7, 2018 at 17:34
• "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. Feb 7, 2018 at 17:35
• @amrsa Yes, that is the question. Feb 7, 2018 at 18:05
• @Yanko The set of terms is {1} which is finite, thus countable. Feb 20, 2018 at 21:59

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.

• Yes, I made a more precise statement. Feb 7, 2018 at 17:36

"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.

• 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 $f^{-1}(1)$? Feb 7, 2018 at 18:10
• @wsaleem The bijection need not be given by the sequence itself ... Feb 7, 2018 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. Feb 7, 2018 at 18:17
• Wsaleem. I honestly fail to see your point. Feb 8, 2018 at 0:25
• f (1)=13,f(2)=8;f (3)=1;f (4)=21;f (5)=3 etc. Is a bijection. Feb 8, 2018 at 0:31

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.