Number of monotonic set functions from all the subsets of some finite set to 0 or 1 Let $N=\{1,2,\ldots n\}$ be some finite set.
Let $f:P(N)\rightarrow\{0,1\}$ be a function such that $A\subset B\rightarrow f(A)\leq f(B)$
I'm trying to find an upper bound to the number of such functions.
Furthermore, I'll call $i,j$ symmetric in $f$ if $\forall A\in P(N)$ with $i,j\not\in A$, $f(A\cup\{i\})=f(A\cup\{j\})$. $f$ is symmetric if there is at least one pair that is symmetric in $f$.
Is there a lower bound on the number of symmetric monotonic functions?
 A: Your first question, about the number of monotonic functions $f:P([n])\to\{0,1\}$ where $[n]=\{1,2,\dots,n\}$, is a classic known as Dedekind's problem.
Your second question is easier. There are $n+2$ symmetric monotonic functions $f:P([n])\to\{0,1\}$, namely, the two constant functions, and for each $k\in[n]$ the function
$$f(X)=\begin{cases}
0\text{ if }|X|\lt k,\\
1\text{ if }|X|\ge k.\\
\end{cases}$$
A: How tight an upper bound do you want/need?
A simple, but clearly in no way strict, upper bound is $2^{(2^n)}$ - i.e. the total number of functions $\mathcal{P}(N) \rightarrow 2$.
However, this upper bound assumes you can choose the values of a function on each element of $\mathcal{P}(N)$ independently, which is certainly not the case with the given condition.
For a slightly better bound;
Consider the chain (under inclusion) $\emptyset, \{1\}, \{1,2\}, ..., N$. 
There are $2^{n+1}$ possible restrictions of a function to this chain, but only $n+2$ of them satisfy $A \subseteq B \Rightarrow f(A) \leq f(B)$.
If we were to then choose the values for the rest of $\mathcal{P}(N)$ as if independently, we now have an upper bound of $(n+2)2^{(2^n - n - 1)}$. This is still not tight, since clearly the function on this chain restricts the allowable values on the rest of $\mathcal{P}(N)$.
For a slightly better bound again;
Consider once more the chain $\emptyset, \{1\}, \{1,2\}, ..., N$. Call this chain $C$.
If $f : \mathcal{P}(N) \rightarrow 2$ satisfies the given condition;
Suppose no member of $C$ has image $1$. Then in particular $N \mapsto 0$, and hence all subsets of $N$ map to $0$, i.e. $f \equiv 0$.
Suppose every member of $C$ has image $1$. Then in particular $\emptyset \mapsto 1$, and hence all supersets of $\emptyset$ map to $1$, i.e. $f \equiv 1$.
Suppose the least member of $C$ that has image $1$ has size $k$, $0 < k < n$. Then this set has $\sum_{i\geq k} {n - k \choose i - k} = 2^{n-k}$ supersets (including itself) which must map to 1, and it's predecessor in $C$ has $2^{k-1}$ subsets (including itself) which must map to 0.
$k$ therefore determines the value of $f$ on $2^{n-k} + 2^{k-1}$ elements of $\mathcal{P}(N)$.
If we allowed the values of the rest of $\mathcal{P}(N)$ to be chosen independently of each other, this would give $2^{2^n - 2^{n-k} - 2^{k-1}}$ functions for a given k.
Summing over possible $k$, we get an upper bound of $2 + \sum_{1 \leq k < n} 2^{2^n - 2^{n-k} - 2^{k-1}}$
(This might simplify, but I haven't looked to closely at the algebra to do so)
Looking in similar ways will give more refinements to the bound at the cost of more complicated reasoning - go as far as you need to go for whatever you want this for, I guess, up to the point of working out exactly how many such functions there are.
