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Proposition The denumerable union of disjoint (in our book, pairwise disjoint) countable nonempty sets are denumerable. (Why add the condition of nonemptiness?)
Proof. 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 $\sqcup S_j$ is denumerable.口
Proposition The denumerable union of countable sets is countable.
Proof. 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 = \sqcup_{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?)口
Hence
Proposition The countable union of countable sets is countable.
Here is a criterion for countability, which might or might not be useful.
Proposition A set $S$ is countable if and only if $S\preceq \mathbb{N}$.
Proof. 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
(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 $S\approx T$.
(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 $S\approx T\approx I_m$.
口
Next, we consider sets with power of continuum. ("The set $A$ has power of continuum" means that $A\approx \mathbb{R}$)
Proposition $\mathbb{R}^n:= \mathbb{R}\times\mathbb{R}\times\cdots\times\mathbb{R}$ for $n$ times, has the power of continuum.
Proof. 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.口
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.
Proposition $[0,1)\approx \wp (\mathbb{N})$ and $\wp (\mathbb{N})\approx \{f\,|\,f:\mathbb{N}\to\{0,1\}\}$
Proof. 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!) Theorem says that $[0,1)\approx \wp (\mathbb{N})$.
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.口
Through the same argument, we derive the following proposition.
Proposition $\wp (\mathbb{A})\approx \{f\,|\,f:\mathbb{A}\to\{0,1\}\}$ for any set $\mathbb{A}$.
Proof. 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.
As an important remark, I have to mention which sets we have compared about their sizes. Indeed, we have the following types of discussion.
(i) $\mathbb{N}\approx \mathbb{Z}\approx \mathbb{Q}\prec \mathbb{R}\approx\mathbb{C}$. (ii) $\mathbb{R}\approx [0,1]\approx [0,1)\approx (0,1)\approx \mathbb{R}^+ \approx [0,\infty)$ (iii) $\wp (\mathbb{N})\approx\mathbb{R} \approx \{f\,|\,f:\mathbb{N}\to \{0,1\}\}$ (iv) $\mathbb{N}\approx \mathbb{N}\times\mathbb{N}\approx\mathbb{N}\times\mathbb{N}\times\mathbb{N}\approx\cdots\approx\mathbb{N}\times\mathbb{N}\times\cdots\mathbb{N}\prec_{(\ast)}\mathbb{N}\times\mathbb{N}\times\cdots$. (v) $\mathbb{R}\approx \mathbb{R}\times\mathbb{R}\approx\mathbb{R}\times\mathbb{R}\times\mathbb{R}\approx\cdots\approx\mathbb{R}\times\mathbb{R}\times\cdots\mathbb{R}\approx_{(\ast\ast)}\mathbb{R}\times\mathbb{R}\times\cdots$.
Note 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.
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}
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 $\mathbb{N}$.口
Next, what are worth thinking for a while are the equinumerosity $(\ast)$ and the "dominance" $(\ast\ast)$. Indeed, imitating Cantor's diagonal method 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.
So we postpone the proof and then show the later assertion $(\ast\ast)$ first.
Proposition $\mathbb{R}\times\mathbb{R}\times\cdots \approx \mathbb{R}$
Proof. 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.口
Proposition $\mathbb{N}\times\mathbb{N}\times\cdots \approx \mathbb{R}$
Proof. 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 $\mathbb{N}\times\mathbb{N}\times\cdots \approx \mathbb{R}$.口
At last, I shall show two essential properties for sets with large cardinality.
Definition $\mathbb{T}$ is defined as all functions from $\mathbb{R}$ to $\mathbb{R}$. $\mathbb{T}_c$ is defined as all functions from $\mathbb{R}$ continuously to $\mathbb{R}$.
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 $\{0,1\}$.
Proposition $\mathbb{R}\prec \mathbb{T}$.
Proof. 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.口
Problem Is $\mathbb{R}\prec \mathbb{T}_c$ or $\mathbb{R}\approx \mathbb{T}_c$ ?? [改自台灣某知名大學碩士班入學考題]
Proposition (Cantor) For every set $A$, $A\prec \wp (A)$, or equivalently, $A\prec \{f\,|\, f:A\to \{0,1\}\}$.
Proof. We prove the later type. Denote $\ddot{A}=\{f\,|\, f:A\to\{0,1\}\}$. Assume $A\approx \ddot{A}$ (Why $A\preceq \ddot{A}$ ??). Let $F:A\to\ddot{A}$ be such a bijection. Define a function $g:A\to \{0,1\}$ such that for $c\in A$, \begin{align} g(c)= \begin{cases} 0, &\text{if $(F(c))(c)=1 $; } \\ 1, &\text{otherwise.} \end{cases} \end{align} We know that $g\in\ddot{A} =F(A)$. For $x\in A$, since $g(x)\ne (F(x))(x)$, we have that $g\ne F(x)$. This means $g\notin F(A)$, a contradiction.口
Solution to Problem. Let $\varphi$ be the mapping: $f\mapsto f|_{\mathbb{Q}}$, the restriction of a given function to the rationals. $\varphi$ is injective because, if $\varphi (f)=\varphi (g)$, i.e. $f|_\mathbb{Q}=g|_\mathbb{Q}$, we're going to show that $f(x)=g(x)$ for all real $x$.
Let $x\in\mathbb{R}\setminus\mathbb{Q}$. Then $x$ can be expressed as a limit of a rational sequence $\{x_j\}_{j=1}^\infty$. Because $f,g$ are both continuous, it follows that \begin{align} f(x)=f(\lim_{j\to\infty}x_j)= \lim_{j\to\infty} f(x_j) = \lim_{j\to\infty} g(x_j)=g(\lim_{j\to\infty}x_j)=g(x)\notag \end{align} Hence $f=g$. Next, denote $\mathbb{Q}$ by \begin{align} \mathbb{Q}=\{q_1, q_2, q_3,\cdots\},\notag \end{align} where $\{q_j\}_{j=1}^\infty$ is a distinct sequence, and then we construct another function $\tau: \{\mathfrak{f}\,|\mathfrak{f}:\mathbb{Q}\to\mathbb{R}\}\to\mathbb{R}\times\mathbb{R}\times\cdots$ by \begin{align} \mathfrak{f}\mapsto (\mathfrak{f}(q_1),\mathfrak{f}(q_2),\mathfrak{f}(q_3)\cdots).\notag \end{align} If $\tau (\mathfrak{f})=\tau(\mathfrak{g})$, i.e. $(\mathfrak{f}(q_1),\mathfrak{f}(q_2),\mathfrak{f}(q_3)\cdots) = (\mathfrak{g}(q_1),\mathfrak{g}(q_2),\mathfrak{g}(q_3)\cdots)$, then for $q\in\mathbb{Q}$, $q=q_k$ for some $k\in\mathbb{N}$. So \begin{align} \mathfrak{f}(q)=\mathfrak{f}(q_k)= \mathfrak{g}(q_k)=\mathfrak{g}(q)\notag \end{align} Hence, $\mathfrak{f}=\mathfrak{g}$.Then $\tau$ is injective.
We find that $\tau\circ\varphi$ is an injection from $\mathbb{T}_c$ to $\mathbb{R}\times\mathbb{R}\times\cdots$, so $\mathbb{T}_c\preceq \mathbb{R}$. Moreover, $\mathbb{R}\preceq\mathbb{T}_c $ because \begin{align} \mathbb{T}_c\supset \bigcup_{r\in\mathbb{R}} \{f:f(x)\equiv r\}.\notag \end{align} Therefore, $\mathbb{R}\approx \mathbb{T}_c$.口
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