For infinite cardinal $\kappa$, $\kappa$ . $\kappa$ = $\kappa$ . Set Theory Enderton p-162, Lemma 6R Enderton's proof goes like below;
Let B be any infinite set of cardinality $\kappa$.
Let $H=${$f|f=$$\emptyset$ or $f: A×A-A$ is a bijection for some A $\subseteq B$}
Then he showed that any chain in H has upper bound in H. Hence H has a maximal element, $f_o$. That's fine I have no problem upto this.
But then he tells that the maximal element $f_o$$:A_o×A_o$ $-A_o$ is a bijection, for some $A_o$ $\subseteq B$. The maximal element $f_o$ might not be a bijection from $B×B$ onto $B$, instead for big enough subset  $A_o$ $\subseteq B$, $f_o$$:A_o×A_o$ $-A_o$ is a bijection. I am unable to comprehend, why should the transfinite process stop at some $A_o$ $\subseteq B$ instead taking whole B. Can someone explain this point with an expository explanation? (Edit: Here I mainly wanted to get an intuition that how the transfinite recursion process will work in general rather than a formal proof)
My guess: For $B=\omega +2$, we might be stop at $A_o=\omega$. But I don't have any insight into the process. Why might we stop at $A_o$.
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There is similar question already here.
show that for an infinite cardinal $k$, $k + k = k$
I understand the Enderton's proof, as well as Asaf's proof. But I am unable to understand the point in Asaf's proof(link of the Asaf's proof given above)/the same point in Enderton's proof that "Why $(K,f)$ might not be a maximal element of the partial order mentioned in the Asaf's proof?"
It would be best, if I asked this in comment below Asaf's answer but I don't have enough reputation to comment. So if someone explain this to me, I will be thankful.
 A: Note: This answer focuses on the sum of cardinals case, the product case is slightly different but the same example works, and is left as an exercise.
Letting $K=\mathbb{N}$. Applying Zorn’s lemma we get the existence of a maximal element under that order, all Zorn’s lemma says is that a maximal element exists and nothing more, the maximal element could for example be $(\mathbb{N},f)$ but it could also be $(\mathbb{N}-\{0\},g)$(Checking this is a maximal element is not hard), so Zorn’s lemma gives no guarantee that the maximal element is $(K,f)$. Luckily, by the way the order is constructed it is not hard to prove that the maximal element must be of the form $(K-F, h)$  where $F$ is some finite set.
Edit: You don't seem to be convinced that $(\mathbb{N}-\{0\}, g)$ where, $g: \mathbb{N}-\{0\} \to 2\times (\mathbb{N}-\{0\})$ is a bijection, is a maximal element. To prove that it is maximal element note that there can not exist a bijection $f:\mathbb{N}\to2\times \mathbb{N}$ extending $g$, because $2\times\mathbb{N}$ has two more elements than $2\times(\mathbb{N}-\{0\}$), while $\mathbb{N}$ has only one more element than $\mathbb{N}-\{0\}$.
A: Given sets $S$ and $T,$ I will denote the class of all functions from $S$ to $T$ by ${}^{S}T.$ I will similarly denote the class of all functions from some subset of $S$ to some subset of $T$ by ${}^{[S]}[T].$ The latter is non-standard notation, as far as I'm aware. It is not difficult to prove that if $S$ and $T$ are sets, then so is ${}^ST,$ and from there, one can prove that ${}^{[S]}[T]$ is likewise a set. Finally, for any set $A,$ I will denote the "successor set" of $A$ by $s(A):=A\cup\{A\}.$
Consider $B=s(\omega):=\omega\cup\{\omega\}.$ Then $$H:=\left\{f\in{}^{[B\times B]}[B]\mid f:A\times A\to A\textrm{ is a bijection for some }A\subseteq B\right\}.$$ Indeed, observing that $\emptyset$ is a function that is readily a bijection $\emptyset\times\emptyset\to\emptyset,$ we have that $\emptyset\in H,$ even though I didn't make it explicit that $\emptyset\in H$ from the definition, the way that Enderton did.
The next thing worth noting is something that you already observed, which is that if $A$ is a finite, non-empty subset of $B,$ then there is no $f\in H$ such that $f:A\times A\to A.$ However, that doesn't mean that $H=\{\emptyset\}.$
Indeed, we can use the following classical construction of a bijection $\omega\times\omega\to\omega.$

*

*Partition $\omega\times\omega$ into sets of ordered pairs as follows. Given $n\in\omega,$ and let $$A_n:=\bigl\{\langle k,m\rangle\in\omega\times\omega\mid k\cup m=n\bigr\}.$$ So, for any element of $A_n,$ we know that neither of its coordinates is greater than $n,$ and at least one of its coordinates is equal to $n.$ It can be shown that the sets $A_n$ are pairwise disjoint, and that their union is all of $\omega\times\omega.$

*It can be shown that for each $n\in\omega,$ $|A_n|=2n+1.$

*Define the relation $\sqsubset$ on $\omega\times\omega$ by $\langle k_1,m_1\rangle\sqsubset\langle k_2,m_2\rangle$ iff either (1) $m_1=m_2$ and $k_1\in k_2$ or (2) $k_1=k_2$ and $m_2\in m_1.$ It can readily be verified that the restriction of $\sqsubset$ to any of the sets $A_n$ is an order relation on $A_n,$ and so a well-order relation (that we might think of as the "clockwise" order).

*Now, we construct a bijection $\omega\times\omega\to\omega$ by recursion, first letting $g(\emptyset,\emptyset)=\emptyset.$ If we have just defined $g(k,\emptyset)$ for some $k\in\omega,$ then we define $fg\bigl(\emptyset,s(k)\bigr):=s\bigl(g(k,\emptyset)\bigr).$ If we have just defined $g(k,m)$ for some $\langle k,m\rangle\in\omega\times\omega$ such that $m\neq\emptyset,$ then letting $n=k\cup m,$ we have that $\langle k,m\rangle\in A_n,$ but is not the $\sqsubset$-greatest element of $A_n,$ so letting $\langle k',m'\rangle$ be the immediate $\sqsubset$-successor of $\langle k,m\rangle$ in $A_n,$ we define $g(k',m'):=s\bigl(g(k,\emptyset)\bigr).$
The gist is that we work our way through all of $\omega\times\omega$ by working through the sets $A_n$ in numerical order, and working through the individual elements of each set $A_n$ in "clockwise order." I leave it to you to verify that the function $g$ thus defined is, in fact, a bijection $\omega\times\omega\to\omega.$ (You can even work out a formula, if you want.)
However, there are several observations about $g$ that are much more relevant!

*

*$g\in H.$

*$g$ is a maximal element of $H$ with respect to set inclusion.

*If $f\in H$ such that $f\subseteq g,$ then either $f=g$ or $f=\emptyset.$

*The largest chain in $H$ including $g$ is $\{\emptyset,g\}.$
Consequently, there need be no transfinite process occurring in $H$ to produce a maximal element at all!
A: I am writing this answer to explain actually what I was asking an expository explanation rather than a formal answer. I thank everyone for engaging and trying to help me.

If we get a bijection from from (B-D)×(B-D) onto (B-D), where cardinality of D is strictly less than that of B, then we can't extend this bijection further to (B-D)$\cup$F × (B-D)$\cup$F onto (B-D)$\cup$F, where cardinality of F strictly less than that of B and F$\cap$D $\neq \phi$. I understand this point from Yoku's comment. I understood the formal proof from Enderton's Set theory(page 162).
What I was trying to understand that in the transfinite construction process within the framework of Zorn's Lemma how might we come to such bijection from (B-D)×(B-D) onto B-D, where $|D|< |B|$ ,$D\subset B$, from starting the construction with some arbitary  bijection from T×T onto T , where |B-T| is not necessarily strictly less |B|. Now  I have spent some time with this problem and  I have an idea how the transfinite proceess will work. I will write an answer highlighting actually what I was expecting.
Suppose $B= \omega$.
Clearly $\phi \in H$. $C=\{ \phi \}$ is chain in $H$. $\cup C \in H$. If this not maximal element of $H$. We can extend this to find a bigger elements. Clearly there is
no bijection from $m×m\to m, m\in \omega$. Hence this construction in $H$ cannot be extended to any finite subset of $\omega$. This construction is different from other usual construction in (transfinite) recursion process where we add some next element and extend the construction.
Suppose $A \subset B$ s.t. $B-A$ is finite.  Then there exists a bijection $t:A×A \to A$. $C=\{ \phi,t \}$ is chain in $H$. $\cup C \in H$. $\phi$ can be extended to $t$. This $t$ cannot be further extended. So $t$ is maximal element of $H$. This one of many possible constructions. Here we started with $\phi$, instead we can start with some arbitary element in $H$. Here we extended $\phi$ to $t:A×A \to A$, where$B-A$ is finite. There are many other   extensions possible for example a bijection  $\{2,4,6,...\}×\{2,4,6,...\} \to \{2,4,6,...\}$. Instead of $\phi$, we can start the construction with a bijection from $P×P\to P$, where $P$ is the set of prime numbers. Different starting point and different extensions will lead us to different constructions. But the point I wanted to understand that every possible construction will lead us to a extension like $t:A×A \to A$ is a bijection, where$B-A$ is finite. Let me explain this below;
Before considering the general case, Let us start the construction with a bijection $h:\{1,3,5,7,..\} × \{1,3,5,7,..\} \to \{1,3,5,7,..\}$. Now if $k:\{2,4,6,..\} × \{2,4,6,..\} \to \{2,4,6,..\}$ is a bijection, then $h\cup k:(\omega -\{0\})× (\omega -\{0\}) \to  (\omega -\{0\})$ is a extension of $h$ in $H$ and is a maximal element of $H$
Let us start with an arbitrary $f: K×K\to K \in H$. Let $T= \omega - K$. Let $P=\{P_0,P_1,P_2,...\}$ be partition of $T$. Let $|P_i|= \aleph_0$, for $i\in v\subset \omega$. Then for each $i\in v$, we a bijection $f_i:P_i × P_i \to P_i$. If $\bigcup_{j\in \omega -v} P_j$ is finite, $f\cup \bigcup_{i\in v} f_i$ will be the maximal element and the construction stop here. Else we will partition $\bigcup_{j\in \omega -v} P_j$ and continue the construction.
