0
$\begingroup$

The notion of sequence is basic notion in combinatorics and can be defined

1.Mapping from a finite set for example $$I_m=\{0,1,2,...,m-1\}$$ to an other set $X$ is finite sequence with terms in $X$ and is denote by $$s=(x_o,x_1,...,x_{m-1}), x_i\in X$$ 2.Mapping from a countable set for example $$\mathbb N=\{0,1,2,...,n,...\}$$ to an other set $X$ is infinite sequence and is denoted by $$s=(x_0,x_1,...,x_n,...), x_i\in X$$

Finite sequnces with terms from a finite set can be enumerated and they we call permutations or variations. Definition of sequences s contain two cases that are extremal.

$$s:\emptyset\to X$$ $$s:X\to\emptyset$$ $X$ is countable

How to deal with such cases. Any suggestions

$\endgroup$
2
  • $\begingroup$ There is no function from a non-empty set into an empty set. Could you explain what your actual question is? Do you want to count empty sub-sequences of a given set? $\endgroup$
    – Phira
    Nov 19 '12 at 12:52
  • $\begingroup$ The case of empty set is very important and must be considered. That maybe explain why for example $0!=1$ $\endgroup$
    – Adi Dani
    Nov 19 '12 at 13:01
5
$\begingroup$

There is exactly one map $s \colon \emptyset \to X$ for any set $X$, it is called the empty map. In the usual interpretation of maps as subsets (here of $\emptyset \times X = \emptyset$) it corresponds to the empty set. In terms of sequences, you can count it as the empty sequence (with zero terms) and denote it $s = ()$. It is for example important as neutral element of juxtraposition in the monoid of all finite sequences in $X$.

On the other side, if $X$ is not empty, then there is no map $X \to \emptyset$. Such a map, given by a set $A \subseteq X \times \emptyset = \emptyset$ must have $$ \forall x \in X \; \exists! y\in \emptyset\; (x,y) \in A = \emptyset. $$ As for non-empty $X$ this can't be true, there is no such map.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.