I'm reading a proof of Tychonoff theorem that using cluster point of a net.
Let $\left\{X_{\alpha}\right\}_{\alpha \in \Lambda}$ be a family of compact topological spaces. Then $X := \prod_{\alpha \in \Lambda} X_\alpha$ together with the product topology is compact.
The proof is given as follows.
Let $\left\{X_{\alpha}\right\}_{\alpha \in \Lambda}$ be a family of compact spaces, let $X=$ $\prod_{\alpha \in \Lambda}$ and let $\left\{f_{d}\right\}_{d \in \mathcal{D}}$ be a net in $X$. As in the proof of Theorem 2.1, we define an element of $\prod_{\alpha \in I} X_{\alpha}$ where $I \subset \Lambda$ to be a partial function. We say that a partial function $f$ is a partial cluster point if it is a cluster point of the net $\left\{\left.f_{d}\right|_{I}\right\}$ for some $I \subset \Lambda$. If there is a partial cluster point with domain $I=\Lambda$, then we have found a cluster point of the net $\left\{f_{d}\right\}$ and proved that $X$ is compact.
Let $P$ be the set of partial cluster points ordered by inclusion. Note that $P \neq \emptyset$ for if we let $\alpha \in \Lambda$ be one index and set $I=\{\alpha\}$, then $\left\{f_{d}(\alpha)\right\}$ is a net in $X_{\alpha}$. Since $X_{\alpha}$ is compact, there is a cluster point $p \in X_{\alpha}$ of the net $\left\{f_{d}(\alpha)\right\}$. Then for $I=\{\alpha\}$, the partial function $f: I \rightarrow X$ defined by $f(\alpha)=p$ is a partial cluster point. Also, every chain in $P$ has an upper bound since the union of the partial cluster points in a chain will also be a partial cluster point. Thus $P$ satisfies the hypotheses of Zorn's Lemma.
Let $g$ be a maximal element of $P$. If the domain $I$ of $g=\Lambda$ then we are done. If the domain of $g \neq \Lambda$, choose an index $\alpha \in \Lambda \backslash I$. Then $\left\{f_{d}(\alpha)\right\}$ is a net in $X_{\alpha}$. Since $X_{\alpha}$ is compact, there is a cluster point $p \in X_{\alpha}$ of the net $\left\{f_{d}(\alpha)\right\}$. Then $h$ defined by
$$
h(\beta)= \begin{cases}p & \text { if } \beta=\alpha \\ g(\beta) & \text { if } \beta \in I\end{cases}
$$
is a partial cluster point with domain $I \cup\{\alpha\}$ extending $g$. This contradicts the maximality of $g$. Therefore, the domain of $g$ is all of $\Lambda$ and the proof is complete.
My questions:
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I could not prove that the union of partial cluster points is again a luster point. This problem is somehow related to below question.
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The maximal element $g$ is a cluster point of $\left\{f_{d} \restriction I\right\}_{d \in \mathcal{D}}$ with $I$ the domain of $g$. However, a tuple of cluster points (w.r.t. the corresponding coordinate) is not necessarily a cluster point in the product topology. A counter-example can be found here. As such, I'm not sure if $h := g \cup \{(\alpha, p)\}$ is a cluster point of $\left\{f_{d}\lceil I \cup\{\alpha\}\}_{d \in \mathcal{D}} .\right.$ Hence we may not have a contradiction as we expect.
Could you elaborate on my confusion?
Update: I added here the proof in my notations.
Let $(E_i)_{i\in I}$ be an arbitrary collection of compact topological spaces. We endow $E:= \prod_{i\in I} E_i$ with the product topology. Then $E$ is compact.
A choice function $f$ is a map from $J \subseteq I$ to $\bigcup_{i\in J} E_i$ such that $f (i) \in E_i$ for all $i \in J$. Let $(f_d)_{d\in D}$ be a net in $E$. Our goal is to find a cluster point of this net. To remove ambiguity, let's recall the definition of a cluster point.
Let $X$ be a topological space and $(x_d)_{d \in D}$ a net in $X$. A point $x \in X$ is a cluster point of $(x_d)_{d \in D}$ if and only if for each neighborhood $U$ of $x$ and each $d \in D$, there is $d' \ge d$ such that $x_{d'} \in U$.
A partial cluster point is a choice function $f:J \to \bigcup_{i\in J} E_i$ such that $f$ is a cluster point of $(f_d \restriction J)_{d\in D}$. Let $\mathcal C$ be the set of all partial cluster points. Clearly, $\mathcal C$ is non-empty. We define a partial order $\le$ on $\mathcal C$ by $$f \le g \iff (\operatorname{dom} f \subseteq \operatorname{dom} g) \text{ and } (g\restriction \operatorname{dom} f = f).$$
Let $(g_t)$ with $g_t: J_t \to \bigcup_{i \in I_t} E_i$ be a chain in $\mathcal C$ and $g := \bigcup g_t$. Let $J := \bigcup J_t$ be the domain of $g$. Next we prove that $g \in \mathcal C$, i.e., $g$ is a cluster point of $(f_d \restriction J)_{d\in D}$. Let $d \in D$ and $U$ be a neighborhood of $g$. There exist open neighborhoods $U_i$ of $g(i)$ in $E_i$ such that $\prod_{i\in J} U_i \subseteq U$. We need to find $d' \ge d$ such that $f_{d'} \restriction J \in U$, or equivalently $f_{d'} (i) \in U_i$ for all $i \in J$. Assume the contrary that such $d'$ does not exist, i.e., $\forall d' \ge d, \exists i\in J, f_{d'}(i) \notin U_i$.
Best Answer
$\newcommand{\dom}{\operatorname{dom}}$You must note that $P$ is ordered by inclusion. Admittedly that's not so clear - how does a function include another? The proof you are following is essentially identical to the proof that I myself saw and took notes on. In that (more clearly written) proof, they partially order $P$ by $g\ge h$ if $\dom g\supseteq\dom h$ and $g\equiv h$ on their common domain.
Then you can hopefully see why the union of an ordered chain is an upper bound for that chain - the domain of the union is indeed a superset of the domains of all the child nets, and the union will agree with the child nets on each of the common domains.
It remains a partial cluster point since any basic neighbourhood of the union will have finite support $F$, and $F$ will be contained in the domain $g_\lambda$ for some $\lambda$ in the chain-index, and $g_\lambda$ is a partial cluster point of $f_\bullet$ so $f_\bullet|_{\dom g_\lambda}$ will be frequently in $F$, hence $f_\bullet|_{\dom\cup g_\lambda}$ will be eventually in $F$.
When I say domain, I mean an index set $I\subset\Lambda$. The proof hinges upon the fact that the largest union of such domains, in the maximal element as guaranteed by Zorn's lemma, must equal $\Lambda$ as otherwise the compactness of all the $X_\alpha$ will imply a contradiction. It is the only time the compactness is used (in the proof I saw).
About the contradiction: since the maximal $g$ is a partial cluster point of $f_\bullet$, there is a subnet $f_{\varphi(\beta)}$ indexed by $\beta\in B$ for which $g$ is a limit point. Since $X_\alpha$ is compact, $f_{\varphi(\beta)}$ is a net which will attain a cluster point $p\in X_\alpha$. Then $h$ is a partial cluster point of the subnet $f_{\varphi(\beta)}$ and by extension a partial cluster point of $f_\bullet$. Note that the counter-example you linked does not apply, since the projection of the subnet $(a_n,b_n)_{n\text{ even}}$ attains the cluster point $0$ in the first coordinate, we note that $0$ is not a cluster point of this subnet in the second coordinate, since the second coordinate will appear $0,2,4,6,\cdots$. This is why this particular choice of $p$ is important.
A more thoroughly written proof can be found here.