Define sequences without sets or natural numbers

Question

I wondered whether it is possible to define a sequence without relying on sets or natural numbers and tried with this definition.

1. A symbol which is not a comma is a sequence.
2. If $S$ is sequence and $s$ a symbol which is not a comma:
   • $S, s$ is a sequence, where every symbol occurring in $S$ precedes $\phi$ and $\phi$ is the last symbol of the sequence.
   • $s, S$ is a sequence, where every symbol occurring in $S$ follows $\phi$ and $\phi$ is the first symbol of the sequence.

So given the string "$a,b,c$", "$a$" is a sequence and "$a,b$" is a sequence too. Therefore "$a,b,c$" is the sequence "$S, c$", where "$S$" is "$a,b$", "$c$" is its last element and "$a,b$" precede it. Similarly "$a$" can be identified as the first element.

Is that correct?

Answer 1

Why not make it even simpler?

1. Any symbol is a sequence
2. Sequences are closed under concatenation

This is similar to the definition of the free monoid.

In terms of addressing elements in a sequence: given the sequence $AsB$, the symbol $s$ is the first element if $A$ is trivial, and the last if $B$ is trivial.

Answer 2

You seem to be attempting a recursive definition of finite sequences. A recursive definition needs 3 things: construction, distinguishing, and exclusion.

For construction:

A symbol which is not a comma is a sequence.

It is very very bad style not to distinguish between a singleton and it's element, between $[a]$ and $a$. This also leaves out the zero length sequence. I suggest simply "$[]$ is a sequence".

If S is sequence and s a symbol which is not a comma...

I suggest just writing "If [a] is a sequence, then [a,x] is a sequence".

Distinguishability:

You can't just assume that because 2 things are written differently that they are different, for example the sets $\{a, b\} = \{b, a\} = \{a, a, b, b\}$. So

$$x \ne y \Rightarrow [x] \ne [y]$$ $$[] \ne [a,\dots]$$ $$[a] \ne [b] \Rightarrow [a,x] \ne [b,x]$$

Exclusion

So far you have said what is a set, but you haven't said what isn't a set. For this induction is the most straightforward, especially since you explicitly don't want to use set theory.