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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 I`m 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.

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You're well on your way.

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 $g$.

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