Showing that $X_K$ is not Connected in the Box Topology

Question

Okay, so I solved this problem discussed here, although I didn't go through trouble of showing $f$ (defined in the link) is a homeomorphism by showing images and preimages of open sets are open; I just used the maps to product theorem. However, I have been wondering whether $X_K$, defined as the consisting of all $(x_i)$ such that $x_i = a_i$ when $i \notin K$, where $K$ is some finite set, is connected in the box topology.

Consider $A = \{(x_i) \in X_K ~|~ x_i \neq a_i ~\forall i \in K \}$ and $B = \{(x_i) \in X_K ~|~ x_i = a_i \mbox{ for some } i \in K \}$. Clearly these are disjoint and clearly $B$ is open in $X_K$: if $U_i = \{a_i\}$ for all $i \notin K$ and $U_i = X_i$ for all $i \in K$ (I suppose $U_i = X_i$ for every $i$ would work, too), then $X_K \cap \prod U_i = B$. Showing $A$ is open, however, is a little different. It seems that I need to choose $U_i = \{a_i\}$ for every $i \notin K$, and $U_i$ an open set not containing $a_i$ for every $i \in K$, in order to show that $X_K \cap \prod U_i = A$. My concern is in choosing $U_i$ so that $a_i$ is not in it. How do I know this is always possible? For what topologically spaces is it possible?

EDIT:

Okay. Here is an attempt at proving $X_K$ is connected when $\prod_{i \in I} X_i$ For simplicity, I will in most places replace "$i \in I$" and "$i \in K$" by "$I$" and "$K$", respectively; for instance, $(x_i)_{i \in I}$ will be replaced by $(x_i)_I$, $\prod_{i \in K} X_i$ by $\prod_K X_i$, etc.

Let $f: X_K \rightarrow \prod_K X_i$ be defined by $f[(x_i)_I] = (x_i)_K$; it is clear that it is a bijective function, so I will proceed to showing it is bicontinuous. Let $\prod_K U_i$ be some basis element in $\prod_K X_i$. I claim that $f^{-1}(\prod_K U_i) = X_K \cap \prod_I V_i$, where $V_i = U_i$ if $i \in K$ and $V_i = X_i$ if $i \notin K$. Suppose $(x_i)_K \in f(\prod_K U_i)$. This happens iff $(x_i)_I \in X_K$ and $f[(x_i)_I] = (x_i)_K \in \prod_K U_i$ iff $x_i \in X_K$ and $x_i \in U_i = V_i$ for $i \in K$ iff $(x_i)_I \in X_k \cap \prod_I V_i$.

Now, let $\prod_I U_i$ be some basis element, and therefore $U_i = \{a_i\}$ if $i \notin K$, and for $i \in K$, $U_i$ is any open set in $X_i$. I claim that $f(\prod_I U_i) = \prod_K U_i$. Let $y \in f(\prod_I U_i)$. This happens iff there exists a $(x_i)_I \in \prod_I U_i$ such that $y = f[(x_i)_I] = (x_i)_K$, where $x_i \in U_i$ for $i \in K$; moreover, this happens if and only if $(x_i)_K \in \prod_K U_i$.

First, is there anything wrong with either proof; any improvement on the ideas or clearer presentation? Second, is there simpler way of doing this?

Answer 1

$X_K$ is just homeomorphic to a finite product of $X_i, i \in K$, and for this finite product the usual and box topology coincide. For the usual product all finite products of connected spaces are connected. So $X_K$ is connected in the box and in the usual topology.

Proof: $p_K: X_K \rightarrow \prod_{ i \in K} X_i$, defined by the projection, is clearly continuous as the restriction of the projection map.If $U = \prod_i U_i \cap X_K$ is basic open on $X_K$ (in the box-subspace topology) ,$p_K[U] = \prod_{i \in K} U_i$ which is open in the finite product topology on the right hand side. So $p_K$ is an open continuous bijection.

BTW, If the $X_i$ are such that each $X_i$ has two disjoint open sets, $\prod_{i \in I} X_i$ is not connected in the box topology. See M.E.Rudin's booklet, or the chapter on box products in the Handbook of Set-Theoretic topology. There are good reasons to prefer the usual product topology (product of compacts are compact, of connected spaces are connected, etc.) over the box topology (where no infinite non-trivial product can be compact or connected).