# Prove for all $f:\mathbb{N}\to\mathbb{N}$ exists $g:\mathbb{N}\to\mathbb{N}$ such that $g$ is immediate successor of $f$ in ${R}$.

We define $$R$$ on $$\mathbb{N}^\mathbb{N}$$ such that: $$\forall f,g:\mathbb{N}\to \mathbb{N}$$, $$fRg$$ if and only if $$\forall n\in \mathbb{N}, f(n)\leq g(n)$$. Then, $$R$$ is partially ordered set on $$\mathbb{N}^\mathbb{N}$$.

Prove or Disprove:

For all $$f:\mathbb{N}\to\mathbb{N}$$ exists $$g:\mathbb{N}\to\mathbb{N}$$ such that $$g$$ is immediate successor of $$f$$ in $${R}$$.

I think this is correct so Im trying to prove it, however I stuck in my proof.

Here is how I tried to prove it:

Let be $$f:\mathbb{N}\to\mathbb{N}$$ a function and we define $$g:\mathbb{N}\to\mathbb{N}$$, $$g(n)=f(n)$$ if n $$\neq$$ 0. and $$g(n)=f(n)+1$$ if n=0.

According to the definition, an immediate successor $$g \in \mathbb{N}^\mathbb{N}$$ of $$f \in \mathbb{N}^\mathbb{N}$$ is such that $$f\neq g$$ and $$fRg$$ and if $$h\in \mathbb{N}^\mathbb{N}$$ satisfies $$fRh$$ and $$hRg$$, then either $$h=f$$ or $$h=g$$.

please let me know if how I start is correct but I don$$$$t have any idea how to continue if that is correct.

Since $$fRh$$ and $$hRg$$, we have for any $$n\in \Bbb N$$ that $$f(n)\leq h(n)\leq g(n)$$. Now, for any $$n\geq 1$$ we have $$f(n) = g(n)$$, which means that we must also have $$h(n) = f(n) = g(n)$$.
Then for $$n = 0$$, we have $$f(n)\leq h(n) \leq g(n) = f(0) + 1$$No matter what $$h(0)$$ is, it must be either $$f(0)$$ or $$g(0)$$, meaning $$h$$ is either the same function as $$f$$ or the same function as $$g$$.