The answer to Q4 is yes. Let $X$ be any infinite set. Wlog $X= Z\times\mathbb N$, where there is a bijection $i:X\to Z\times \lbrace0\rbrace$. For $x=(z,n)$ write $x+1$ for $(x,n+1)$. (z,n+1)$. [Typo corrected.] You are given countably many finitary functions$g_1, g_2, \ldots$. We may assume there is a pairing function$x*y$among them, so we may as well assume that all of them are binary. (Due to Sierpinski, I think. E.g.,$g(x,y,z) = h(x*(y*z)) $for some unary$h$.) Now there is a binary function$f$satisfying the following for all$x,y\in X$: 1.$f(x,x) = x+1$. 2.$f(x, x+1) = i(x)$. 3.$f(i(x)+k,i(y)) = g_k(x,y)$for$k=1,2,\ldots$. Clearly$f$generates the functions$x+1$,$i(x)$, and$g_k$for all$k$. 2 fixed stupid error The answer to Q4 is yes. Let$X$be any infinite set. Wlog$X= Z\times\mathbb N$, where there is a bijection$i:X\to Z\times \lbrace0\rbrace $. For$x=(z,n)$write$x+1$for$(x,n+1)$. You are given countably many finitary functions$g_1, g_2, \ldots$. We may assume there is a pairing function$x*y$among them, so we may as well assume that all of them are binary. (Due to Sierpinski, I think. E.g.,$g(x,y,z) = h(x*(y*z)) $for some unary$h$.) Now there is a binary function$f$satisfying the following for all$x,y\in X$: 1.$f(x,x) = x+1$. 2.$f(x, x+1) = i(x)$. 3.$f(i(x), y+k) f(i(x)+k,i(y)) = g_k(x,y)$for$k=1,2,\ldots$. Clearly$f$generates the functions$x+1$,$i(x)$, and$g_k$for all$k$. 1 The answer to Q4 is yes. Let$X$be any infinite set. Wlog$X= Z\times\mathbb N$, where there is a bijection$i:X\to Z\times \lbrace0\rbrace $. For$x=(z,n)$write$x+1$for$(x,n+1)$. You are given countably many finitary functions$g_1, g_2, \ldots$. We may assume there is a pairing function$x*y$among them, so we may as well assume that all of them are binary. (Due to Sierpinski, I think. E.g.,$g(x,y,z) = h(x*(y*z)) $for some unary$h$.) Now there is a binary function$f$satisfying the following for all$x,y\in X$: 1.$f(x,x) = x+1$. 2.$f(x, x+1) = i(x)$. 3.$f(i(x), y+k) = g_k(x,y)$for$k=1,2,\ldots$. Clearly$f$generates the functions$x+1$,$i(x)$, and$g_k$for all$k\$.