

<b><i>Proposition</i></b>&nbsp;&nbsp; The denumerable union of disjoint (in our book, pairwise disjoint) countable <u>nonempty</u> sets are denumerable. (Why add the condition of nonemptiness?)<br>
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<i>Proof.</i> Let $S_1$, $S_2$, ... be all the given sets. Then we know that each $S_j$ is in the form
\begin{align}
S_j=\{s_{j1},s_{j2},\cdots s_{j,n}\}\,\,\,\,\, or\,\,\,\,\, S_j=\{s_{j1}, s_{j2}\cdots \}\notag
\end{align}
Then the mapping $s_{jk}\mapsto \langle j,k\rangle$, where $k$ is in the "domain" of $S_j$, is bijective. Hence its image must be a subset of $\mathbb{N}\times\mathbb{N}$. Moreover, $\{s_{11},s_{21}, s_{31}\cdots\}$ forms a denumerable set. Hence a previous example yields that $\coprod S_j$ is denumerable. 
<p align=right>f</p>
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<b><i>Proposition</i></b>&nbsp;&nbsp; The denumerable union of countable sets is countable.<br>
<br>
<br>
<i>Proof.</i> Let $S_1$, $S_2$, ... be all given sets. Set
\begin{align}
T_1&=S_1\notag\\
T_{j+1}&=S_{j+1}\setminus T_j \qquad\qquad \textit{for }j\in\mathbb{N}\notag
\end{align}
Then $\{T_j\}_{j=1}^\infty$ is pairwise disjoint. So

\begin{align}
S=\bigcup_{j=1}^\infty S_j = \coprod_{j=1}^\infty T_j\notag
\end{align}

is a denumerable union of disjoint countable sets. It'll be countable. (When is it non-denumerable? Can it even be empty?)

<p align=right>f</p>


&nbsp;&nbsp;&nbsp;&nbsp;Hence<br>
<br>
<b><i>Proposition</i></b>&nbsp;&nbsp; The countable union of countable sets is countable. <br>
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Here is a criterion for countability, which might or might not be useful. <br>
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<br>
<b><i>Proposition</i></b>&nbsp;&nbsp; A set $S$ is countable if and only if $S\preceq \mathbb{N}$.<br>
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<br>
<i>Proof.</i> If $S$ is denumerable, then $S\approx \mathbb{N}$ and hence $S\preceq \mathbb{N}$. If $S$ if finite, then $S\preceq\mathbb{N}$ because $I_n\preceq \mathbb{N}$. Conversly, Let $f$ be one-to-one from $S$ into $\mathbb{N}$. Let $T=Im(f)$. Then<br>
<br>
(a) If $T$ contains no maximum, then $\mathbb{N}\preceq T$ (why?). Since $T\subset \mathbb{N}$, we obtain that $\mathbb{N}\approx T$. Since $f$ is onto $T$, it follows that $S\approx T$. <br>
(b) If $T$ contains a maximum, say $n_0\in\mathbb{N}$. Then $T\subset I_{n_0}$. Hence $T$ is equinumerous to some $I_m$ (why? This might be proved by induction.). Similarly, the assertion is proved by $S\approx T\approx I_m$. <br>

<p align=right>f</p>

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Next, we consider sets with power of continuum. ("The set $A$ has <b>power of continuum</b>" means that $A\approx \mathbb{R}$)<br>
<br>
<b><i>Proposition</i></b>&nbsp;&nbsp; $\mathbb{R}^n:= \mathbb{R}\times\mathbb{R}\times\cdots\times\mathbb{R}$ for $n$ times, has the power of continuum. <br>
<br>
<br>
<i>Proof.</i> Denote

\begin{align}
\mathfrak{m}=\{n\,\,:\,\,\mathbb{R}^n\,\,\textit{has the power of continuum}   \}\notag
\end{align}

It's shown that $1,2\in\mathfrak{m}$. Suppose that $k\in\mathfrak{m}$. Let $f:\mathbb{R}\to\mathbb{R}^k$ and $G:\mathbb{R}\to\mathbb{R}\times \mathbb{R}$ be bijections. Write

\begin{align}
f(x)&=(f_1 (x), f_2 (x)\cdots f_k (x))\notag\\
g(x)&=(g_1 (x), g_2 (x))\notag
\end{align}

Then define $h:\mathbb{R}\to\mathbb{R}^{k+1}$ as 
\begin{align}
h(x)=(f_1\circ g_1 (x), f_2 \circ g_1 (x)\cdots f_k \circ g_1 (x), g_2 (x)),\notag
\end{align}

$h(x)$ is then one-to-one and onto. 
<p align=right>f</p>
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What is the connection between the natural number set $\mathbb{N}$ and the real number set $\mathbb{R}$ in the sense of cardinality (the "number" of elements in a set)?? In fact, a thoroughly obvervation would tell us the following proposition. <br>
<br>
<b><i>Proposition</i></b>&nbsp;&nbsp; $[0,1)\approx \wp (\mathbb{N})$ and $\wp (\mathbb{N})\approx \{f\,|\,f:\mathbb{N}\to\{0,1\}\}$<br>
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<i>Proof.</i> Define the mapping $f$ by $0.a_1 a_2\cdots\mapsto\{j\,|\,a_j=1\}$, and the mapping $g$ by $B\subset\mathbb{N}\mapsto 0.b_1 b_2 b_3\cdots$ in binary where $b_{2j}=1$ if and only if $j\in B$ while all other decimal digits are 0. Hence $f$ and $g$ play roles as injections in both directions. Schr$\ddot{o}$der Bernstein (Be careful for the spelling<i>!</i>) Theorem says that $[0,1)\approx \wp (\mathbb{N})$. <br>
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The other assertion is similar. Take $\varphi$ as the mapping who maps a subset $B$ of $\mathbb{N}$ to the function $f:\mathbb{N}\to\{0,1\}$ such that $f(n)=1$ if and only if $n\in B$. Another  mapping $\tau$ is chosen to map a function $f:\mathbb{N}\to\{0,1\}$ to the subset $B=\{j\in\mathbb{N}\,:\,f(j)=1\}$. They both are injection (Verify it!). Therefore Schr$\ddot{o}$der Bernstein leads us to the conclusion. 
<p align=right>f</p>
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Through the same argument, we derive the following proposition. <br>
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<b><i>Proposition</i></b>&nbsp;&nbsp; $\wp (\mathbb{A})\approx \{f\,|\,f:\mathbb{A}\to\{0,1\}\}$ for any set $\mathbb{A}$. <br>
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<i>Proof.</i> Define the mapping $\varphi$ who maps a subset $A$ of $\mathbb{A}$ to the function $f:\mathbb{A}\to\{0,1\}$ such that $f(a)=1$ if and only if $a\in A$. Another  mapping $\tau$ is chosen to map a function $f:\mathbb{A}\to\{0,1\}$ to the subset $A=\{a\in\mathbb{A}\,:\,f(a)=1\}$. They both are injection and then we prove the equinumerosity. 





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As an important remark, I have to mention which sets we have compared about their sizes. Indeed, we have the following types of discussion.<br>
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<blockquote><blockquote>
<table border=0>
<tr>
<td align=right>(i)</td> <td>&nbsp;&nbsp; $\mathbb{N}\approx \mathbb{Z}\approx \mathbb{Q}\prec \mathbb{R}\approx\mathbb{C}$.</td>
</tr>

<tr>
<td align=right>(ii)</td> <td>&nbsp;&nbsp; $\mathbb{R}\approx [0,1]\approx [0,1)\approx (0,1)\approx \mathbb{R}^+ \approx [0,\infty)$ </td>
</tr>

<tr>
<td align=right>(iii)</td> <td>&nbsp;&nbsp; $\wp (\mathbb{N})\approx\mathbb{R} \approx \{f\,|\,f:\mathbb{N}\to \{0,1\}\}$</td>
</tr>

<tr>
<td align=right>(iv)</td> <td>&nbsp;&nbsp; $\mathbb{N}\approx \mathbb{N}\times\mathbb{N}\approx\mathbb{N}\times\mathbb{N}\times\mathbb{N}\approx\cdots\approx\mathbb{N}\times\mathbb{N}\times\cdots\mathbb{N}\prec_{(\ast)}\mathbb{N}\times\mathbb{N}\times\cdots$.</td>
</tr>

<tr>
<td align=right>(v)</td> <td>&nbsp;&nbsp;$\mathbb{R}\approx \mathbb{R}\times\mathbb{R}\approx\mathbb{R}\times\mathbb{R}\times\mathbb{R}\approx\cdots\approx\mathbb{R}\times\mathbb{R}\times\cdots\mathbb{R}\approx_{(\ast\ast)}\mathbb{R}\times\mathbb{R}\times\cdots$.</td>

</table>
</blockquote></blockquote>



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<u> Note</u>&nbsp;&nbsp;&nbsp; I have to emphasize the meaning of the notation $\mathbb{A}\times\mathbb{A}\times\cdots$ for a given set $\mathbb{A}$. Recall that we use Order Pairs as elements in the product sets of given sets, namely
\begin{align}
\prod_{j=1}^n A_j = A_1\times A_2 \times\cdots\times A_n = \{(a_1, a_2,\cdots a_n)\,:\,a_j\in A_j  \},\notag
\end{align}
and an interesting notion  for an order pair is that, we might view it as a function ! By this I mean, after we define
\begin{align}
(a_1,a_2,\cdots a_n)=\{  \langle 1,a_1\rangle, \langle 2,a_2\rangle\cdots \langle n,a_n\rangle  \}\notag
\end{align}
the concept about an order pair is illustrated. <br>
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The notation $\mathbb{A}\times\mathbb{A}\times\cdots$ seems to express all "infinite order pairs". Comparing with above, if  $(a_1, \cdots , a_n)$  stands for a finite sequence, it is natural to convent that $(a_1,a_2,\cdots)$ means an infinite sequence, i.e. 
\begin{align}
(a_1,a_2,\cdots)=\{\langle 1,a_1\rangle , \langle 2,a_2\rangle\cdots   \}\notag
\end{align}
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Directly thinking, we finally define  $\mathbb{A}\times \mathbb{A}\times \cdots = \{a\,\,|\,\, a:\mathbb{N}\to\mathbb{A}\}$. In particular, if we choose $\mathbb{A}=\mathbb{N}$, then $\mathbb{N}\times\mathbb{N}\times\cdots = \{a\,\,|\,\, a:\mathbb{N}\to\mathbb{A}\}$, all sequences in $\mathbb{N}$. 
<p align=right>f</p>
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Next, what are worth thinking for a while are the equinumerosity $(\ast)$ and the "dominance" $(\ast\ast)$. Indeed, imitating <i><u>Cantor's diagonal method</u></i> for real numbers, we can show that $\mathbb{N}\times\mathbb{N}\cdots$ is uncountable. However, we still have no idea about whether it has power of continuum. <br>
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So we postpone the proof and then show the later assertion $(\ast\ast)$ first. <br>
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<b><i>Proposition</i></b>&nbsp;&nbsp; $\mathbb{R}\times\mathbb{R}\times\cdots \approx \mathbb{R}$<br>
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<i>Proof.</i> We ought to define an  injection $\varphi :[0,1)\times [0,1)\times \cdots \to [0,1)$ throughout decimal expressions. For $r= (r_1, r_2, r_3, \cdots)\in\mathbb{R}\times\mathbb{R}\times\cdots$, write
\begin{align}
r_1 &= 0.r_{11} r_{12} r_{13} r_{14}\cdots\notag\\
r_2 &= 0.r_{21} r_{22} r_{23} r_{24}\cdots\notag\\
r_3 &= 0.r_{31} r_{32} r_{33} r_{34}\cdots\notag\\
r_4 &= 0.r_{41} r_{42} r_{43} r_{44}\cdots\notag\\
&\vdots\notag
\end{align}
Then we define $\varphi (r) = 0.r_{11} r_{21} r_{22} r_{12} r_{13} r_{23} r_{33} r_{32} r_{31} r_{41} r_{42} r_{43} r_{44} r_{34}\cdots$. We know $\varphi$ is injective (Show!), and we can easily find a converse injection. Hence Schr$\ddot{o}$der Bernstein Theorem concludes the equinumerosity. 
<p align=right>f</p>
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<b><i>Proposition</i></b>&nbsp;&nbsp; $\mathbb{N}\times\mathbb{N}\times\cdots \approx \mathbb{R}$<br>
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<i>Proof.</i> Since 
\begin{align}
\mathbb{R}\approx \{0,1\}\times\{0,1\}\times\cdots \preceq \mathbb{N}\times\mathbb{N}\times\cdots \preceq \mathbb{R}\times\mathbb{R}\times\cdots\approx \mathbb{R},\notag
\end{align}
immediately we have $\mathbb{N}\times\mathbb{N}\times\cdots \approx \mathbb{R}$. 
<p align=right>f</p>

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At last, I shall show two essential properties for sets with large cardinality. <br>
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<b><i>Definition</i></b>&nbsp;&nbsp; $\mathbb{T}$ is defined as all functions from $\mathbb{R}$ to $\mathbb{R}$. $\mathbb{T}_c$ is defined as all functions from $\mathbb{R}$ continuously to $\mathbb{R}$. <br>
<font color=#ffffff>In all my articles, $\mathbb{T}$ is defined as all functions from $\mathbb{R}$ to $\mathbb{R}$; $\mathbb{T}_2$ is defined as all functions from $\mathbb{R}$ to $\{0,1\}$. </font><br>
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<b><i>Proposition</i></b>&nbsp;&nbsp; $\mathbb{R}\prec \mathbb{T}$.<br>
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<i>Proof.</i> Assume $\mathbb{R}\approx \mathbb{T}$ (Why $\mathbb{R}\preceq \mathbb{T}$ ??). Let $F:\mathbb{R}\to\mathbb{T}$ be such a bijection. Define a function $g:\mathbb{R}\to \mathbb{R}$ such that for $c\in \mathbb{R}$, 
\begin{align}
 g(c)=
  \begin{cases}
    0,  &\text{if $(F(c))(c)=1 $; } \\
    1,    &\text{otherwise.}
  \end{cases}
\end{align}

(This is the "diagonal argument" on continuum.) We know that $g\in\mathbb{T} =F(\mathbb{R})$. For $x\in \mathbb{R}$, since $g(x)\ne (F(x))(x)$, we have that $g\ne F(x)$. This means $g\notin F(\mathbb{R})$, a contradiction.
<p align=right>f</p>
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<b><i>Problem</i></b>&nbsp;&nbsp; Is $\mathbb{R}\prec \mathbb{T}_c$ or $\mathbb{R}\approx \mathbb{T}_c$ ??    [ۥxWYWjǺӤhZJǦD]<br>
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<font color=#0000ff><b><i>Proposition</i>&nbsp;(Cantor)</b>&nbsp;&nbsp; For every set $A$, $A\prec \wp (A)$, or equivalently, $A\prec \{f\,|\, f:A\to \{0,1\}\}$. <br>
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<i>Proof.</i> We prove the later type. Denote $\ddot{\mathbb{A}}=\{f\,|\, f:\mathbb{A}\to\{0,1\}\}$. Assume $\mathbb{A}\approx \ddot{\mathbb{A}}$ (Why $\mathbb{A}\preceq \ddot{\mathbb{A}}$ ??). Let $F:\mathbb{A}\to\ddot{\mathbb{A}}$ be such a bijection. Define a function $g:\mathbb{A}\to \{0,1\}$ such that for $c\in \mathbb{A}$, 
\begin{align}
 g(c)=
  \begin{cases}
    0,  &\text{if $(F(c))(c)=1 $; } \\
    1,    &\text{otherwise.}
  \end{cases}
\end{align}

 We know that $g\in\ddot{\mathbb{A}} =F(\mathbb{A})$. For $x\in \mathbb{A}$, since $g(x)\ne (F(x))(x)$, we have that $g\ne F(x)$. This means $g\notin F(\mathbb{A})$, a contradiction.
<p align=right>f</p>
</font>



