A proof that the operation of concatenation has an identity element

Question

I'm relatively new to proofing and am wondering if this is an acceptable proof. The book for anyone who would like to reference it is "A Book of Abstract Algebra" by Charles Pinter. It is problem $3$ of problem set D in chapter $2$.

In short, the question defines the operation of concatenation on $A^*$, the set of all sequences of symbols in the alphabet $A$ as "say $a=a_1a_2\cdots a_n$ and $b=b_1b_2\cdots b_m$, then $a\cdot b=a_1a_2\cdots a_nb_1b_2\cdots b_m$." The question then mentions that the the empty sequence is denoted by say $\emptyset$. The problem then asks us to prove that there is an identity element for the operation of $A^*$. My proof is as follows

Theorem: The operation of concatenation denoted by $\cdot$ and defined by $a\cdot b=a_1a_2...a_n b_1b_2...b_n$ on $A^*$, the set of all sequences of symbols in the alphabet $A$, has an identity element.

Proof: Suppose $x\in A^*$.

Suppose $\emptyset$ denotes the empty sequence.

$\emptyset\cdot x=x_1x_2...x_n$ and $x\cdot\emptyset=x_1x_2...x_n$,

Thus, by the definition of an identity element, $\emptyset$ is the identity element of $\cdot$ on $A^*$.

Therefore there is an identity element for the operation of concatenation on the set of all sequences in the alphabet $A$.

My largest concern is that $\emptyset\cdot x=x_1x_2...x_n$ and $x\cdot\emptyset=x_1x_2...x_n$, does not look like $\forall x\in A^*~~~ \emptyset \cdot x=x$ and $x\cdot \emptyset=x$ if that makes any sense.

Answer 1

The proof is fine as it is written. You could use an extra line immediately following $x\in A^*$ where you state:

Let $x\in A^*$ where $x=x_1x_2\dots x_n$ with each $x_i\in A$.

This will clear any confusion as to what you are referring to when you refer to $x\cdot \emptyset = x_1x_2\dots x_n$ later on since previously the $x_1x_2\dots x_n$ were not defined or mentioned.

The fact that you did not specify anything about $x$ itself apart from that it was an element of $A^*$ implies that $x$ could have been anything. Since the result you showed follows regardless what $x$ is, it must have been true for every $x$.

You did somewhat tacitly assume that $x$ is of length $n\geq 3$ in the way that you formatted your proof (since you wrote it as $x_1x_2\cdots x_n$), but it is easily forgivable and ignorable in this case. It might not hurt to mention briefly as a side note why the proof obviously works if $x$ were a length 0,1, or 2 string.