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Here is Prob. 10, Sec. 10, in the book Topology by James R. Munkres, 2nd edition:

Theorem. Let $J$ and $C$ be well-ordered sets; assume that there is no surjective function mapping a section of $J$ onto $C$. Then there exists a unique function $h \colon J \to C$ satisfying the equation $$ h(x) = \mathrm{smallest} \left[ C - h \left( S_x \right) \right] \tag{*} $$ for each $x \in J$, where $S_x$ is the section of $J$ by $x$.

Proof.

(a) If $h$ and $k$ map sections of $J$, or all of $J$, into $C$ and satisfy (*) for all $x$ in their respective domains, show that $h(x) = k(x)$ for all $x$ in both domains.

(b) If there exists a function $h \colon S_\alpha \to C$ satisfying (*), show that there exists a function $k \colon S_\alpha \cup \{ \alpha \} \to C$ satisfying (*).

(c) If $K \subset J$ and for all $\alpha \in K$ there exists a function $h_\alpha \colon S_\alpha \to C$ satisfying (*), show that there exists a function $$ k \colon \bigcup_{ \alpha \in K} S_\alpha \to C $$ satisfying (*).

(d) Show by transfinite induction that for every $\beta \in J$, there exists a function $h_\beta \colon S_\beta \to C$ satisfying (*). [Hint: If $\beta$ has an immediate predecessor $\alpha$, then $S_\beta = S_\alpha \cup \{ \alpha \}$. If not, $S_\beta$ is the union of all $S_\alpha$ with $\alpha < \beta$. ]

(e) Prove the theorem.

My Attempt:

Part (a):

Suppose that there exists an element $x \in \mathrm{Dom} (h) \cap \mathrm{Dom} (k) $ such that $h(x) \neq k(x)$. Then the set of all such $x$ is a non-empty subset of $J$ and hence has a smallest element $\alpha$, say.

Since each of the functions $h$ and $k$ is defined either on all of $J$ or on some section of $J$ and since $\alpha$ is in both domains, therefore we can conclude that both $h$ and $k$ are defined on the section $S_\alpha$ of $J$ by $\alpha$. [Am I right?]

And since $\alpha$ is the smallest element common to the domains of $h$ and $k$ at which these two functions differ, we can also conclude that $h(x) = k(x)$ for all $x \in S_\alpha$, and therefore $$ h \left( S_\alpha \right) = k \left( S_\alpha \right). \tag{1} $$ Now as both $h$ and $k$ satisfy (*), so we can conclude from (1) that $$ h( \alpha ) = \mathrm{smallest} \left[ C - h \left( S_\alpha \right) \right] = \mathrm{smallest} \left[ C - k \left( S_\alpha \right) \right] = k( \alpha), $$ which contradicts our choice of $\alpha$. Hence the functions $h$ and $k$ coincide on the common part of their domains, as required.

Is this proof correct and clear enough? If not, then where are the problems in it?

Part (b):

Suppose that, for some $\alpha \in J$, there exists a function $h \colon S_\alpha \to C$ satisfying (*). Then, according to the hypothesis of this theorem, this function $h$ cannot be surjective. So the set $h \left( S_\alpha \right)$ is a proper subset of $C$, and therefore the set $C - h \left( S_\alpha \right)$, being a non-empty subset of the well-ordered set $C$, must have a smallest element.

Now let the function $k \colon S_\alpha \cup \{ \alpha \} \to C$ be defined as follows: $$ k(x) = \begin{cases} h(x) \ & \mbox{ if } x \in S_\alpha, \\ \mathrm{smallest} \left[ C - h \left( S_\alpha \right) \right] \ & \mbox{ if } x = \alpha. \end{cases} $$ Then $k$ also satisfies (*). [Am I right?]

Is this proof correct and clear enough too? If not, then where are the issues?

Part (c):

Suppsoe $K \subset J$ and suppose that for each $\alpha \in K$ there exists a function $h_\alpha \colon S_\alpha \to C$ satisfying (*). We assume that our set $K$ is non-empty.

Let $x$ be an arbitrary element of $\bigcup_{\alpha \in K} S_\alpha$. Then there exists at least one $\alpha \in K$ such that $x \in S_\alpha$.

Suppoes that $\alpha$ and $\beta$ are any two distinct elements of $K$ such that $x \in S_\alpha \cap S_\beta$. As $\alpha, \beta \in J$ and as $J$ is a simply ordered set, so either $\alpha < \beta$ or $\beta < \alpha$. We assume without loss of generality that $\alpha < \beta$. Then $S_\alpha \cap S_\beta = S_\alpha$. So by Part (a) we can conclude that $$ h_\beta (x) = h_\alpha(x). \tag{2} $$ Now let $K_x$ be the subset of $K$ consisting of all the elements $\alpha \in K$ such that $x \in S_\alpha$; that is, let $$ K_x \colon= \left\{ \ \alpha \in K \ \colon \ x \in S_\alpha \ \right\}. \tag{3} $$ Then $K_x$, being a non-empty subset of $K$ and hence of $J$, has a smallest element $\alpha_x$, say. Hence using the same reasoning as the one used in arriving at (2) we can conclude that $$ h_\alpha(x) = h_{\alpha_x}(x) \ \mbox{ for all } \alpha \in K_x. \tag{4} $$

Now let $k \colon \bigcup_{ \alpha \in K} S_\alpha \to C$ be the function defined by the formula $$ k(x) = h_{\alpha_x}(x) \ \mbox{ for all } x \in \bigcup_{\alpha \in K} S_\alpha, \tag{5} $$ where $\alpha_x$ is the smallest element of the set $K_x$ in (3) above.

Now we show that this function $k$ also satisfies (*).

For any point $x \in \bigcup_{\alpha \in K} S_\alpha$, we note that, if $\alpha_x$ is the smallest element in $K$ such that $x \in S_{\alpha_x}$, then $k(x) = h_{\alpha_x}(x)$.

Let us fix an element $x \in \bigcup_{\alpha \in K} S_\alpha$. Now let us take any element $v \in S_{x}$. Then $v \in J$ and $v < x < \alpha_x$, and we also have $\alpha_v \leq \alpha_x$, where $\alpha_v$ is the smallest element of the set $K_v$ as defined in (3) above, which in turn implies that $S_{\alpha_v} \subset S_{\alpha_x}$ and so using (a) we can also conclude that $$ h_{\alpha_x}(v) = h_{\alpha_v}(v) = k(v). $$ That is, $$ k(v) = h_{\alpha_x}(v) \ \mbox{ for all } v \in S_{x}. $$ Therefore we can conclude that $$ k \left( S_{x} \right) = h_{\alpha_x} \left( S_{x} \right), \tag{6} $$ and hence $$ k(x) = h_{\alpha_x}(x) = \mathrm{smallest} \left[ C - h_{\alpha_x} \left( S_x \right) \right] = \mathrm{smallest} \left[ C - k \left( S_x \right) \right]. $$ Thus our function $k$ also satisfies (*).

Is this proof correct? If so, then is it clear enough too? If not, then where are the problems in it in either logic or clarity?

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  • $\begingroup$ What you did looks good to me! $\endgroup$ – mathcounterexamples.net Jan 2 '18 at 19:39
  • $\begingroup$ @mathcounterexamples.net can you please carefully check the very last paragraph in the proof of Part (c), beginning with the words "Now we show that this function $k$ also satisfies (*)"? I'm wondering if the argument can be improved here. Can you please examine the very last equality in my proof again? $\endgroup$ – Saaqib Mahmood Mar 2 '18 at 8:31
  • $\begingroup$ @mathcounterexamples.net I would be really grateful if you could take time going through my post, especially my proof of Part (c). I'm sure I've improved it quite a bit. What are your comments? $\endgroup$ – Saaqib Mahmood Mar 4 '18 at 12:05

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