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Let $\leq$ be a partial order relation in $\Bbb{N^N}$ set definied:

$f\leq g$ if and only if for all $n\in\Bbb N$, $f(n)\leq g(n)$

bottom cone of function $f$ we call the set $D_f = \{g\in\Bbb{N^N}\mid g \leq f\}$

  1. Indicate all cardinal numbers that are powers of bottom cone.
  2. What kind of power is a set of functions for which bottom cone is infinite?
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up vote 3 down vote accepted

What sort of $f$ would have the property that $D_f$ is finite? It means that there is only a finite number of functions which are less or equal than $f$ at each point. For example if $f(n)=0$ for all $n$, then $D_f=\{f\}$. If $f(0)=1$ and $f(n)=0$ for $n>0$, then there are only two possible functions which are in $D_f$.

It follows that if $D_f$ is finite then $f(n)=0$ for all but finitely many $n$'s. But how many functions can be in $D_f$? Well if $f(0)=k$ and $f(n)=0$ otherwise how much functions can be in $D_f$? Only $k$. But this is true for every $k$, so $D_f$ can be finite with any possible finite value, except zero.

Next let's see what happens if $D_f$ is infinite. In such case we can immediately say that $S_f=\{n\in\mathbb N\mid f(n)\neq 0\}$ is an infinite set. But this means that $S_f$ has cardinality $\aleph_0$ and it has $2^{\aleph_0}$ subsets. Let $A\subseteq S_f$, then the indicator function $\chi_A(n)=0$ for $n\notin A$ and $\chi_A(n)=1$ for $n\in A$, is such that $\chi_A\in D_f$.

We found an injection from a set of cardinality $2^{\aleph_0}$ into $D_f$ just from the assumption that $S_f$ is infinite. Because there are only $2^{\aleph_0}$ functions in $\mathbb{N^N}$ it follows that $D_f$ can be either finite or it has size $2^{\aleph_0}$.

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Great, thank You! :-) –  Steve Jan 6 '13 at 14:20
    
I've forgot, what will be the power of set of functions, for which the bottom cone is infinite? –  Steve Jan 6 '13 at 14:21
    
@test: Use the last result to show that for every infinite $A$, subset of $\mathbb N$, the set $D_{\chi_A}$ is infinite. –  Asaf Karagila Jan 6 '13 at 14:23
    
thanks one more! –  Steve Jan 6 '13 at 14:38
    
One thing - We can't achieve $D_f$ with zero elements, right? –  Steve Jan 6 '13 at 14:53
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