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What is the cardinal number of these sets :

$1.$ The set of finite sequences that its items are taken from inifinite sequences of $\mathbb Z.$

$2.$ The set of infinite sequences that its items are taken from finite sequences of $\mathbb R.$

$3.$ $A=\left\{f: \mathbb R_+ \to \mathbb R_+ | f(x)\leq e^{-x} \right\} $

$4.$ $C:=\left\{h \in \left\{ 0,1 \right\}^\mathbb N| \frac{1}{n+1} \sum_{k=0}^{n} h(k) \leq \frac{1}{2} \right\} $

My solution:

$1.$ I know that |Infinite sequence($\mathbb Z$)|=$|\mathbb Z^\mathbb N| = \mathfrak{c}$ but i dont know how to deal with finite sequence of infinite sequence($\mathbb Z$).

$2.$ I know |inifnite sequence of finite sequence($(\mathbb R ^n)^ \mathbb N$)|=|$\mathbb R^n|^{|(\mathbb N)|}=|\mathbb R^{|(\mathbb N)|}|=\mathfrak{c}$.

$3.$ I have no idea how to deal with this except High limit is $|\mathbb R|^{|\mathbb R|} = 2^{\mathfrak{c}} $

$4.$ High limit is $\mathfrak{c}$ but i cant find low limit.

Is my proof correct for 2 correct ? Please help in Exercises 2,3,4.

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  • $\begingroup$ for 3 the bounds don't matter since they contain functions from $\mathbb{R}\rightarrow (0,1)$ the result is $\beth_2=2^{2^{\aleph_0}}$ not the cardinality of the continuum. For 4) consider the sequences $s$ which are equal to zero on all $n$ that aren't a positive power of $2$ $\endgroup$
    – MIO
    Commented Sep 10, 2021 at 19:50
  • $\begingroup$ @aldodecristo i was wrong , $|\mathbb R|^{|\mathbb R|}=2^\mathfrak{c}$ , fixed it. But i dont understand how to prove $|\mathbb{R}\rightarrow (0,1)|=|A|$ because there are inifnity functions that cant be fit to A image , like $f(x)=e^{(\frac{-x}{2})} $ am i wrong ? $\endgroup$
    – Algo
    Commented Sep 10, 2021 at 20:23
  • $\begingroup$ @RobArthan Sorry, fixed it. $\endgroup$
    – Algo
    Commented Sep 10, 2021 at 20:32
  • $\begingroup$ oh sorry it's bounded by $e^{-x}$ but what you can do is show that there is a bijection from the functions that are bounded by $e^{-x}$ and the ones that are bounded by $1$. let $f$ be bounded by $e^{-x}$ then $\frac{f(x)}{e^{-x}}$ is bounded by $1$ try and show that the map $f\mapsto \frac{f(x)}{e^{-x}}$ is a bijection $\endgroup$
    – MIO
    Commented Sep 11, 2021 at 11:55
  • $\begingroup$ What do you mean by "finite sequence of infinite sequence on $\Bbb Z$". Besides being bad English, there are many ways it might be interpreted, and some of those have different cardinalities than others. And similarly for your second question. Please be specific about what sets you are referring to. $\endgroup$ Commented Sep 11, 2021 at 13:18

1 Answer 1

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I don't know your conventions, but I'll explain using $0\in \Bbb N, 0 \notin \Bbb R_+$ and $N_n := \{k \in \Bbb N\mid k < n\}$ (other conventions do not change the answer, but modify the explanation).

  1. The set of all infinite sequences with values in $\Bbb Z$ is indeed $\Bbb Z^{\Bbb N}$. For a fixed $n$, the set of all sequences of length $n$ into $\Bbb Z^{\Bbb N}$ is $\left(\Bbb Z^{\Bbb N}\right)^{N_n}$. So this set is $$S =\bigcup_{n=0}^\infty \left(\Bbb Z^{\Bbb N}\right)^{N_n}$$Since the sets being unioned are disjoint, we have $$|S| = \sum_{n=0}^\infty \left|\left(\Bbb Z^{\Bbb N}\right)^{N_n}\right| = \sum_{n=0}^\infty \left|\Bbb Z^{\Bbb N}\right|^{|N_n|} = \sum_{n=0}^\infty \mathfrak c^n$$Do you know what $\mathfrak c^n$ is for finite $n$?
  2. Your analysis shows what happens for a fixed $n$. But the set description includes all finite sequences, regardless of length. This should not be too hard to fix.
  3. $f(x) \mapsto e^xf(x)$ provides a bijection between $A$ and the set $A_1$ of all functions $f : \Bbb R_+ \to (0,1)$. And $\tan\left(\frac\pi2 x\right)$ provides a bijection between $(0,1)$ and $\Bbb R_+$.
  4. $C$ is the set of sequences $h$ into $\{0,1\}$ such that for each $n, h$ takes on at least as many $0$ values as $1$ values at locations $\le n$. Consider an arbitrary sequence into $\{0,1\}$. We can build a new sequence by inserting an extra $0$ in front of each $1$ in the original sequence. This ensures that the resulting sequence will be in $C$. The process is also injective. You can convert the output sequence back into the original by removing any $0$ immediately preceding a $1$. So there is an injection from $\{0,1\}^{\Bbb N}$ into $C$.
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  • $\begingroup$ Thank you ! About excersies 2 , $S =\bigcup_{n=0}^\infty \Bbb R^{{N_n}^{\mathbb N}} \implies\sum_{n=0}^\infty |\Bbb R|^{{|N_n|}^{|\mathbb N|}}= |\Bbb R|^{|\mathbb N|} = \mathfrak c $ am i right ? $\endgroup$
    – Algo
    Commented Sep 13, 2021 at 21:05
  • $\begingroup$ Did you notice in my explanation of (1) how much care I took to make sure the order of exponentiation was clear? You really need to take the same care. Unlike addition and multiplication, exponentation is not associative. In general, it matters in which order you do the exponentiations. And in your calculation, you did them in the wrong order. $\endgroup$ Commented Sep 14, 2021 at 0:01
  • $\begingroup$ $S =\bigcup_{n=0}^\infty \Bbb R^{{N_n}^{\mathbb N}} \implies(\sum_{n=0}^\infty |{\Bbb R^{{N_n}}|)^{|\mathbb N|}}= (\sum_{n=0}^\infty {|\Bbb R|^{|{N_n}|})^{|\mathbb N|}} = \mathfrak c^{\aleph_0}= \mathfrak c$ , I am not sure i am right. $\endgroup$
    – Algo
    Commented Sep 14, 2021 at 7:56
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    $\begingroup$ I suggest breaking it down a little further: $$\sum_n |\Bbb R|^{|N_n|} = \sum_n {\mathfrak c}^n = \sum_n \mathfrak c = \aleph_0 \cdot \mathfrak c = \mathfrak c$$ But it is correct. I put "in general" in my last comment exactly because it made no difference in this case. Infinite cardinals have a trivial arithmetric, so calculation errors often make no difference. $\endgroup$ Commented Sep 14, 2021 at 12:14

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