Let $\mathcal F$ be the set of mappings $f: \Bbb N \to \Bbb N$ for which $f(m) \le f(n)$ for $m \le n$. Show that $\mathcal F$ is uncountable

Question

Does my attempt contain gap? Any suggestion is appreciated! Thank you so much!

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Let $\mathcal F$ be the set of mappings $f: \Bbb N \to \Bbb N$ for which $f(m) \le f(n)$ for $m \le n$. Show that $\mathcal F$ is uncountable.

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My attempt:

Let $\mathcal F^\star$ be the set of strictly increasing mappings from $\Bbb N$ to $\Bbb N$. It's clear that $\mathcal F^\star \subseteq \mathcal F$ and thus $|\mathcal F^\star| \le |\mathcal F|$.

If $f\in \mathcal F^\star$, then $\operatorname{ran}f$ contains at least $2$ elements. Thus $|\mathcal{P}(\Bbb N)| \le |\mathcal F^\star|$.

Hence $|\mathcal{P}(\Bbb N)| \le |\mathcal F^\star| \le |\mathcal F|$. We know that $\mathcal{P}(\Bbb N)$ is uncountable. It follows that $\mathcal F$ is uncountable.

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Update: From users' comment and answer, I fixed my proof here.

We know that the set of bounded subsets of $\Bbb N$ is countable and $\mathcal{P}(\Bbb N)$ is uncountable. Thus the set of unbounded subsets of $\Bbb N$ is uncountable.

Furthermore, each unbounded subset of $\Bbb N$ corresponds to a unique mapping $f\in \mathcal F^\star$. Thus $\mathcal F^\star$ is uncountable. Since $|\mathcal F^\star| \le |\mathcal F|$, $\mathcal F$ is uncountable.

Answer 1

I'm not sure how you conclude that $|\mathcal P(\mathbb N)|\le |\mathcal F^\star|$.

Let $g:\mathbb N \to \{ 0,1\}$ be an element of $\{0,1\}^\mathbb N$. Consider $f_g:\mathbb N\to\mathbb N$ defined by $$ f_g(n) = \sum_{k=0}^ng(k)$$ Note that $f_g\in\mathcal F$ for every $g\in\{0,1\}^\mathbb N$, and $$f_g=f_{g'}\iff g=g',$$ thus $g\mapsto f_g$ is an injection $\{0,1\}^{\mathbb N}\to\mathcal F$.

Answer 2

For each not empty A in P(N) find an increasing
f$_A:$N -> N in F with range f$_A$ = A.

Thus there is a surjection h:F -> P(N) - {empty set}
and F is uncountable.