1
$\begingroup$

Let $S = \lbrace 1, 2, 3, . . . , n\rbrace$ be a set. Consider a function $f : S \to S$. A subset $D$ of $S$ is said to be invariant if $\forall$ x $\in$ $D$ we have $f(x) \in D$. The empty set and $S$ are also considered as invariant subsets. By $\deg(f)$ we define the number of invariant subsets $D$ of $S$ for the function $f$.

The questions are--

  1. Show that there exists a function $f : S \to S$ such that $\deg(f) = 2$.

  2. Show that for every $1 \leq k \leq n$ there exists a function $f : S \to S$ such that $\deg(f) = 2^k$.

For the first case I tried to make guess a function such that $f(x)=x+1$ for all $x\in[1,n-1]$ and $f(n)=1$. Clearly the set $S$ satisfies the condition.But I could not find out another set which satisfies this function. Any help is appreciated. Thanks.

1 Answers 1

1

For the first question, define $$f:S \rightarrow S: x \mapsto \begin{cases}x+1 &\text{if }x \neq n \\ 1 &\text{if }x=n \end{cases}$$ In cycle notation this is $(1\, 2\, 3\cdots n)$. Suppose $D$ a non-empty set of $S$ such that $f(D) \subseteq D$. Take any element $m \in D$. Since $f$ cycles through whole $S$ we get that $D=S$. This means that the only invariant subsets are $\emptyset$ and $S$.

For the second question define $f:S \rightarrow S$ bijective such that $f$ has exactly $k$ cycles ($1$-cycles included). The $i$th cycle corresponds to an invariant subset $D_i$ such that $S$ is the union of $D_1, \ldots, D_k$. The $D_i$'s are pairwise disjoint. It can be proven that each invariant subset is the union of some $D_i$'s. Hence you get $2^k$ invariant subsets, thus $\deg(f)=2^k$.