3
$\begingroup$

Show that if $(X_i)_{i \in \mathcal I}$ where $X_i$ is a topological space for every $i \in \mathcal I$, then $X_i$ is connected for every $i$ if and only if $\prod_{i \in \mathcal I} X_i$ is connected.

I could do one implication: If I define the projection map $\Pi_i:\prod_{i \in \mathcal I} X_i \to X_i$, then since $\Pi_i$ is continuous, it follows $X_i$ is connected for every $i \in \mathcal I$.

I need help to prove that if each $X_i$ is connected, then $\prod_{i \in \mathcal I} X_i$ is connected. Thanks in advance.

$\endgroup$
  • $\begingroup$ What can you do if you suppose that $\prod X_i$ is not connected? $\endgroup$ – Travis Willse Aug 27 '14 at 15:27
5
$\begingroup$

First prove that any finite product of connected spaces is connected. This can be done by induction on the number of spaces, and for two connected spaces $X$ and $Y$ we see that $X \times Y$ is connected, by observing that $X \times Y = (\cup_{x \in X} (\{x\} \times Y)) \cup X \times \{y_0\}$, where $y_0 \in Y$ is some fixed point. This is connected because every set $\{x\} \times Y$ is homeomorphic to $Y$ (hence connected) and each of them intersects $X \times \{y_0\}$ (in $(x,y_0)$), which is also connected, as it is homeomorphic to $X$. So the union is connected by standard theorems on unions of connected sets. Now finish the induction. So for every finite set $I$, $\prod_{i \in I} X_i$ is connected.

Now if $I$ is infinite, fix points $p_i \in X_i$ for each $i$, and define $$Y = \{ (x_i) \in \prod_i X_i: \{i \in I: x_i \neq p_i \} \text{ is finite }\}$$

Now for each fixed finite subset $F \subset I$, define $Y_F = \{ (x_i) \in \prod_i X_i: \forall i \notin F: x_i = p_i \}$. By the obvious homeomorphism, $Y_F$ is homeomorphic to $\prod_{i \in F} X_i$, which is connected by the first paragraph. So all $Y_F$ are connected, all contain the point $(p_i)_{i \in I}$ of $\prod_{i \in I} X_i$, and their union (over all finite subsets $F$ of $I$) equals $Y$. So again by standard theorems on the union of connected subsets of a space, $Y$ is a connected subspace of $\prod_{i \in I} X_i$.

Finally note that $Y$ is dense in $\prod_{i \in I} X_i$, because every basic open subset $O$ of the product depends on a finite subset of $I$, in the sense that $O = \prod_{i \in I} U_i$ where all $U_i \subset X_i$ are non-empty open and there is some finite subset $F \subset I$ such that $U_i = X_i$ for all $i \notin F$. Pick $q_i \in U_i$ for $i \in F$ and set $q_i = p_i $ for $i \notin F$. The $(q_i)_{i \in I}$ is in $O \cap Y_F \subset O \cap Y$, so every (basic) open subset of $\prod_{i \in I} X_i$ intersects $Y$.

Now use that the closure of a connected set is connected to conclude that $\prod_{i \in I} X_i$ is connected, also for infinite $I$.

$\endgroup$
  • $\begingroup$ Why the union of $Y_F$ is equal to $Y$ ? Could you explain please ? $\endgroup$ – user486983 Feb 25 '18 at 6:15
  • $\begingroup$ @bella It's by definition: Let $y \in Y$. Then define $F_y = \{i \in I: y_i \neq p_i \}$ which is finite by definition of $y$ being in $Y$. Then again by definition, $y \in Y_{F_i}$. So $y \in \bigcup \{Y_F: F \subseteq I \text { finite }\}$. $Y$ is sometimes called the $\sigma$-product w.r.t. $(p_i)_i$, BTW. $\endgroup$ – Henno Brandsma Feb 25 '18 at 7:17
  • $\begingroup$ Henno my brain is going to explote $\endgroup$ – user486983 Feb 25 '18 at 7:36
  • $\begingroup$ What is $Y_{F_i}$? $\endgroup$ – user486983 Feb 25 '18 at 7:37
  • $\begingroup$ @bella I meant $Y_F$ for $F = F_y$, not $F_i$. so $Y_{F_y}$, really. typo. $\endgroup$ – Henno Brandsma Feb 25 '18 at 8:17
2
$\begingroup$

Hint: Suppose that $F: \prod X_i \to \{0,1\}$ is continuous. For some choice of $x_j \in X_j$ for all $j \neq i$, we can define a continuous function from $X_i \to \{0,1\}$ by $$ f(y) = F\left(\prod_{j \in I} y_j\right) $$ Where $y_j = x_j$ when $j \neq i$ and $y_i = y$.

Now, if $X_i$ is connected, then $F$ must be constant on every set of the form $f(X_i)$ with $f,F$ as described above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.