Free monoid of natural numbers excluding zero

Question

Let $U$ be a set of finite sequences like $\{1,1\cdot1,1\cdot2,\dots,1\cdot 2\cdot3,\dots\}$, i.e. there is no $0$ element in any sequence and all sequences start from $1$.

Can this set be defined as free monoid? (is this set a free monoid?)

I have tried the following:

Let $\mathcal{N}_{>0}$ be the set of positive integers. Let $U$ be the free monoid with identity $0$, generated by $\mathcal{N}_{>0}$ with operation $\cdot$

but after this discussion (see comments below the question) I am confused, is it properly or not?

Answer 1

Let $S$ be the set of (nonempty) sequences that start with $1$ and then all the following terms are greater than $1$. Then $U$ is freely generated by $S$ as a monoid. Indeed, every element $x\in U$ can be written as a composition of a sequence elements of $S$ in a unique way: just break $x$ up into subsequences of consecutive terms, each one starting with a $1$ and going until either you reach the next $1$ in $x$ or the end of $x$.

Answer 2

More generally, consider an alphabet $A$ (in your case the set of positive integers). The free monoid on $A$ (usually denoted $A^*$) is the set of words on $A$, equipped with the concatenation product. For instance, if $A = \{a, b, c\}$ the product of the words $abac$ and $caba$ is the word $abaccaba$. The empty word is the neutral element of the free monoid $A^*$.

Now, let $a$ be an element of $A$ (in your case $a = 1$) and let $B = A - \{a\}$. Then the set $S = aB^*$ is a suffix code, which means that no word of $S$ is a suffix of another element of $S$. In this case, the submonoid of $A^*$ generated by $S$ is a free monoid.