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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 .

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