Is my proof correct? Any general feedback on proof writing style is also much appreciated!
1. Theorem
The infinite product of connected spaces is connected in the product topology. In particular, let $ \{X_\alpha \}_{\alpha \in A}$ be an infinite collection of connected topological spaces. Then
$$ X := \prod_{\alpha \in A} X_\alpha $$
is connected, where $ (X, \tau_X) $ has the product topology.
2. Lemma
Fix $ x \in X $ and define $X_K \subseteq X$ for finite index set $K$ as
$$ X_K := \{ z \in X : \forall i \notin K, z_i = x_i \} $$
Then $x \in X_K$ and $ X_K $ is homeomorphic to $ \prod_{k \in K} X_k $.
3. Proof of Lemma
Define $ f : X_K \rightarrow \prod X_k $ as simply removing coordinates not in $K$.
This is clearly surjective. And injective: if $f(u) = 0$, then $u_k = 0$ for each $ k \in K $ (and the coordinates outside of $K$ are fixed).
Next, note that for any open $U \subset X_K$, as subspace of $X$, there exists in $V \in \tau_X$ such that $ U = V \cap X_K $. Thus, each $U_k$ is open in $X_k$ and the remaining coordinates are fixed.
Since any open $ W \subset \prod X_k $ has open $W_k$ in $X_k$, it's immediate that $f$ and $f^{-1}$ map $ U_k \leftrightarrow U_k $. It follows that f is a homeomorphism. $\square$
4. Proof of Theorem
Again fix $ x \in X $. Let $ E := C(x) $, the connected component of $x$ in $X$.
Since E is closed (we will not prove this here), it suffices to show that E is dense.
First note that $\prod X_K$ is connected for any for any finite $K$, since the finite product of connected sets is connected (not proven here). Now we use the Lemma to conclude that $X_K = f^{-1}(\prod X_k)$ is connected (connectedness is a topological property). It follows that $X_K \subseteq E $ for all finite $K$.
It remains to show that $ F:= \bigcup \{ X_K : finite~K \} $ is dense in X. This will follow from a property of the product topology. Also, note that the arbitrary union of connected sets is connected (not shown here), so $F \subseteq E$.
We will show that all points of $X$ are adherent to $X_K$ for some finite $K$ and then be done. Fix $y \in X$ and a y-neighborhood $U \in \tau_X$. From the product topology, we have
$ U = \prod U_\alpha $ with $U_\alpha \ne X_\alpha $ only finitely often with $ U_\alpha$ open in $X_\alpha$.
Then let $I$ be the finite set of indices where $U_\alpha \ne X_\alpha$.
Then we can choose a $z \in X_K$ with $K = I$ and $ z_k = k_k $ for each $ k \in K$.
It follows that $z \in U$ and all points of $X$ are adherent to $F \subseteq E$. $\square$
5. Related posts.
This post does not provide a complete proof but is relevant.
I think this post provides a complete proof. Thanks @Henno Brandsma.
6. Updates
Thanks to @diracdeltafunk and @Henno Brandsma for pointing out errors/confusing conventions in my original proof.
I think this version (2022/2/12 at 12:30 ET) fixes those issues, so I'll treat this as an answer unless someone raises another issue.
Best Answer
I'm submitting a finalized version of my original proof as a proposed answer.
1. Theorem
The infinite product of connected spaces is connected in the product topology. In particular, let $ \{X_\alpha \}_{\alpha \in A}$ be an infinite collection of connected topological spaces. Then $$ X := \prod_{\alpha \in A} X_\alpha $$ is connected, where $ (X, \tau_X) $ has the product topology.
2. Lemma
Fix $ x \in X $ and define $X_K \subseteq X$ for finite index set $K$ as
$$ X_K := \{ z \in X : \forall i \notin K, z_i = x_i \} $$
Then $x \in X_K$ and $ X_K $ is homeomorphic to $ \prod_{k \in K} X_k $.
3. Proof of Lemma
Define $ f : X_K \rightarrow \prod X_k $ as simply removing coordinates not in $K$.
This is clearly surjective. And injective: if $f(u) = 0$, then $u_k = 0$ for each $ k \in K $ (and the coordinates outside of $K$ are fixed).
Next, note that for any open $U \subset X_K$, as subspace of $X$, there exists in $V \in \tau_X$ such that $ U = V \cap X_K $. Thus, each $U_k$ is open in $X_k$ and the remaining coordinates are fixed.
Since any open $ W \subset \prod X_k $ has open $W_k$ in $X_k$, it's immediate that $f$ and $f^{-1}$ map $ U_k \leftrightarrow U_k $. It follows that f is a homeomorphism. $\square$
4. Proof of Theorem
Again fix $ x \in X $. Let $ E := C(x) $, the connected component of $x$ in $X$.
Since E is closed (we will not prove this here), it suffices to show that E is dense.
First note that $\prod X_K$ is connected for any for any finite $K$, since the finite product of connected sets is connected (not proven here). Now we use the Lemma to conclude that $X_K = f^{-1}(\prod X_k)$ is connected (connectedness is a topological property). It follows that $X_K \subseteq E $ for all finite $K$.
It remains to show that $ F:= \bigcup \{ X_K : finite~K \} $ is dense in X. This will follow from a property of the product topology. Also, note that the arbitrary union of connected sets is connected (not shown here), so $F \subseteq E$.
We will show that all points of $X$ are adherent to $X_K$ for some finite $K$ and then be done. Fix $y \in X$ and a y-neighborhood $U \in \tau_X$. From the product topology, we have $ U = \prod U_\alpha $ with $U_\alpha \ne X_\alpha $ only finitely often with $ U_\alpha$ open in $X_\alpha$.
Then let $I$ be the finite set of indices where $U_\alpha \ne X_\alpha$. Then we can choose a $z \in X_K$ with $K = I$ and $ z_k = k_k $ for each $ k \in K$.
It follows that $z \in U$ and all points of $X$ are adherent to $F \subseteq E$. $\square$