The cardinal number different fields function

Question

What is the cardinal number of these sets :

1. $A=\left\{f \in \mathbb N^\mathbb N : \sum_{n\in\mathbb N}f(n)<\infty\right\}$

2. $B=\left\{ f \in \mathbb R^\mathbb N : |\sum\limits_{k=0}^m f(k)|\leq1 | for \space all \space m \in \mathbb N \right\}$

3. $C=\left\{ f \in \mathbb Z^\mathbb N : |\sum\limits_{k=0}^m f(k)|\leq1 | for \space all \space m \in \mathbb N \right\}$

4. $D=\left\{ f \in \mathbb (0,1)^\mathbb N : | f(x)\lt f(x+1)\right\}$

5. True\False. If A,B are inifnity sets and $|A|>|B|$ so $|A|^{|B|}=|A|.$

My solution :

1. Since the sum is finite so each function (a sequence) must be finite. The sum of finite sequences over $\mathbb N$ is $\aleph_0$.

2. The high limit: $|\mathbb R ^ \mathbb N| = 2^{\aleph_0} $ and its clear that $C \subset B$ so $|B|=2^{\aleph_0}.$

3. High limit : $|\mathbb Z ^ \mathbb N| = 2^{\aleph_0} $ , since $\left\{0,1\right\}^\mathbb N \subset C $ and $|\left\{0,1\right\}^\mathbb N| = 2^{\aleph_0} $ so $|C|=2^{\aleph_0}.$

4. Hight limit : $|(0,1)^\mathbb N|=2^{\aleph_0}.$ I haven no idea how to find the low limit.

5. I think its true but i dont know how to prove it. (Maybe induction ?).

Is my proofs correct ?

Answer 1

1. Ok
2. Ok assuming 3 is ok, which it isn't.
3. Not ok because $\left\{0,1\right\}^\mathbb N \not\subseteq C$. However Let $g(m)=\sum\limits_{k=0}^m f(k)$ then $g\in\{-1,0,1\}^\mathbb{N}$ and you can go back from $g$-s to $f$-s by letting $f(m)=g(m)-g(m-1)$.
4. Let $f(n)=1-1/n$ for $n\ge 2$, $f(1)=1/4$ and $f(0)\in (0,1/4)$.
5. $\aleph_\omega>\aleph_0$ and $\aleph_\omega^{\aleph_0}>\aleph_\omega$ by Konig's theorem.