Let $\lbrace X_\alpha \rbrace_{\alpha \in J}$ be an indexed family of connected spaces; let $X$ be the product space $$X=\prod_{\alpha \in J} X_\alpha$$
Let a$=(a_\alpha)$ be a fixed point of $X$.
Given any finite subset $K$ of $J$, let $X_K$ denote the subspace of $X$ consisting of all points x$=(x_\alpha)$ such that $x_\alpha=a_\alpha$ for $\alpha \notin K$. Show that $X_K$ is connected.
As the title suggests, this is part 1 of 3 to show that $X$ is connected (found in Munkres' Topology, second edition, p. 152). The other parts look fairly straightforward, but this one has been a little tricky for me; the main problem is that $J$ might be uncountable, which makes traditional index notation unusable.
My approach so far has been to show that $X_K$ is homeomorphic to a finite sub-product of $X$. If $|K|=n$ (which is known to be finite), then let $\alpha_1,…,\alpha_n$ be the $n$ indices for which x and a differ. Also let $$Z=\prod_{m=\alpha_1}^{\alpha_n} X_m$$
I suspect that $X_K$ is homeomorphic to $Z$. The most natural map would be $f: X_K \rightarrow Z$ defined by, for x $\in X_K$, "stripping off" all coordinates of x which disagree with a, leaving only finitely many. The inverse would take z $\in Z$ and "fill in" the coordinates of a that are missing.
I haven't tried showing that $f$ is invertible or continuous for uncountable $J$, but I have tried for countable indices; invertibility doesn't seem to be a problem, but continuity is tripping me up:
When $J$ is countable, without loss of generality, take $J=\mathbb{N}$ and $K=\lbrace 1,2,…,n \rbrace$, so $$Z=\prod_{m=1}^n X_m $$Now a=$(a_1,a_2,…)$, and x $\in X_K$ has the form $(x_1,…,x_n,a_{n+1},…)$. Thus $f(x_1,…,x_n,a_{n+1},…)=(x_1,…,x_n)$.
Suppose $U$ is open in $Z$. Then $U=U_1 \times … \times U_n$, where $U_i$ is open in $X_i$. $f^{-1}(U)=U_1 \times … \times U_n \times \lbrace a_{n+1} \rbrace \times…$.
It seems difficult to conclude that $f^{-1}(U)$ is open, as the singleton sets appended on the end may or may not be open in their individual sets, depending on the properties of the $X_i$. The approach also seems awkward when attempting to generalize to uncountable indices.
Apologies for the long question statement, but one more request – please NO FULL SOLUTIONS. I am working through this for an assignment and would like to figure it out on my own, even though I am momentarily stumped.
Best Answer
Remember that you don't need to show $f^{-1}(U)$ is open in $X$, only that it is open in the subspace $X_K$ of $X$. I would also remark that the uncountable case should be no harder than the countable case; it's just a matter of notation. If $J$ is uncountable, you can still label the elements of $K$ with natural numbers $\{1,2,\dots,n\}$; the only difference is that the remaining elements of $K$ will not be other natural numbers (and you don't really care what they are).