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## Infinite cardinal numbers.

Prove that if $x$ is an infinite cardinal number, then $x+x=x$ and $x\cdot x=x$.

Solution: Let $D=A\times \{0,1\}=(A\times \{0\})\cup (A\times \{1\})$. Then $card D=card A+card A=x+x$.

We denote $\mathcal{P}^*(A)=\mathcal{P}(A)\setminus \{\emptyset\}$ the set containing non-void subsets of $A$. For any non-void set $X \subseteq A$ we denote $\mathcal{F}(X)=\{f:X \to X \times \{0,1\}:\ f\ \textrm{ is bijective} \}$.

Let $\displaystyle \mathcal{F}=\bigcup_{X\in \mathcal{P}^*(A)}\mathcal{F}(X)$. The set $A$, being infinite, has a infinite countable subset; so there exists $h:\mathbb{N} \to A$ injective. Then the set $E=h(\mathbb{N})\subseteq A$ is countable because $h:\mathbb{N} \to E$ is bijective and $card E=\aleph_0$. Furthermore we have $card (E\times \{0,1\})=card ((E\times \{0\})\cup(E \times\{1\}))=card (E\times \{0\}) +card (E \times \{1\})=\aleph_0+\aleph_0=\aleph_0=card E$. Therefore we have a function $f:E \to E\times \{0,1\}$ bijective. This shows that the set $\mathcal{F}$ if non-void.

Consider the binary relation $\mathcal{R}=\{(f,g) \in \mathcal F\times \mathcal F:\ g \textrm{ is an extension of } f\}.$
( The function $g:Z\to T$ is an extension of $f:X \to Y$ if $X \subseteq A$ and $g(\alpha)=f(\alpha)\ \forall \alpha \in X$)

We prove that (\mathcal F,\mathcal R) \$ is a partially ordered set (i.e. $\mathcal R$ order relation).
$(f,f) \in \mathcal R$: $f$ is one of its extensions because $Dom\ f\subseteq Dom\ f$ and $f(x)=f(x)\ \forall x \in Dom\ f$.
$(f,g),\ (g,f) \in \mathcal R \Rightarrow f=g$: We have $f:M\to N$ and $g:P\to Q$. Because $(f,g),\ (g,f) \in \mathcal R$ we must have $M\subseteq P$ and $P \subseteq M$, so $M=P$ and $f(\alpha)=g(\alpha)\ \forall \alpha \in Dom\ f=M=P=Dom\ g$. Therefore $f=g$.
$(f,g),\ (g,h) \in \mathcal R\Rightarrow (f,h) \in \mathcal R$: We have $f:F_1 \to F_2,\ g:G_1 \to G_2,\ h:H_1\to H_2$. Furthermore, $F_1 \subseteq G_1 \subseteq H_1$ and $h(\alpha)=g(\alpha),\ \forall \alpha \in G_1$ and $g(\alpha)=f(\alpha),\ \forall \alpha \in F_1$. Take $\alpha \in F_1$. We have $\alpha \in F_1 \subseteq G_1$. So $f(\alpha)=g(\alpha)=h(\alpha)$. This means that $h$ is an extension of $f$. Therefore $(f,h) \in \mathcal R$.

We have proved that $\mathcal R$ is an order on $\mathcal F$.

Take $\mathcal L=(f_\alpha)_{\alpha \in J} \subset \mathcal F$ a totally ordered set. It’s obvious that for any $\alpha \in J$ The function $f_\alpha :X_\alpha \to X_\alpha \times \{0,1\}$ is bijective. Even more, for $\alpha,\ \beta \in J$ the function $f_\alpha:X_\alpha \to X_\alpha \times \{0,1\}$ is an extension for $f_\beta:X_\beta\to X_\beta\times\{0,1\}$ or $f_\beta:X_\beta\to X_\beta\times\{0,1\}$ is an extension for $f_\alpha:X_\alpha \to X_\alpha \times \{0,1\}$.

From above, we get that if $X_0=\displaystyle \bigcup_{\alpha \in J}X_\alpha$, then the function $f_0:X_0\to X_0\times \{0,1\}$ defined for any $y \in X_0$ by $f_0(y)=f_\alpha(y)$, for all $\alpha \in J$ with $y \in X_\alpha$, is bijective.

$f_0$ is a bijective extension for all $f_\alpha \in \mathcal L$. Therefore $f_0$ is an upper bound for $\mathcal L$. Since $\mathcal L$ was chosen randomly, we proved that any totally ordered set has an upper bound.

Using Zorn‘s lemma, we find that $(\mathcal F,\mathcal R)$ has at least one maximal element $f^*:X^* \to X^*\times \{0,1\}$. This means that $X^*$ is non-void and $f^*:X^* \to X^* \times \{0,1\}$ cannot be extended

We’ll show that $Y=A \setminus X^*$ is finite. Suppose that $Y$ is infinite. then we can find an injection $v:\mathbb{N} \to Y$. Therefore, $H=v(\mathbb{N})$ and $H \times \{0,1\}$ are countably infinite, so we can find a bijection $g^*:H \to H \times \{0,1\}$. Let $f^{**}:X^*\cup H \to (X^*\cup H)\times \{0,1\}$ defined for all $y \in X^*\cup H$ by
$f^{**}(y)= \begin{cases} f^*(y), y \in X^* \\ g^*(y), y \in H\end{cases}$.

Then $f^{**}:X^*\cup H \to (X^*\cup H)\times \{0,1\}$ is in $\mathcal F$ and extends $f*:X^* \to X^* \times \{0,1\}$. This contradicts the maximality of $f^*:X^* \to X^*\times\{0,1\}$.

Therefore, $Y=A\setminus X^{*}$ is finite and $X^*=A\setminus Y$ is infinite. From here we get the equality $card A=card (X^*\cup Y)=card X^*=card (X^*\times \{0,1\})=$
$=card ((X\times \{0\})\cup (X\times \{1\}))=card (X^*\times\{0\})+card (X* \times \{1\})=$
$=card X^*+cardX^*=card A+card A$. Therefore $x+x=x$, where $x=card A$.