Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm working on the "iff"-relation given by:

$X=\prod_{i\in I}X_i$ is connected iff each $X_i$ non-empty is connected for all $i\in I$.

I could prove the "$\Rightarrow$"-direction very easyly. I also proved that a finite product of connected spaces is connected. Now i want to prove the following:

  • Choose $z=(z_i)\in\prod_{i\in I}X_i$. For every finite subset $J\subset I$ is the set $X_J:=\left\{x\in X:x_i=z_i\ \forall I-J\right\}$ connected.

I have given the follwoing proof: This set is homeomorphic with a finite product $\prod_{j\in J}X_j$ given by the map defined by: $x=(x_j)_{j\in J}$ mapped on $y=(y_i)_{i\in I}$ such that $y_j=x_j$ if $j\in J$ and $y_j=z_j$ if $j\notin J$. This mapping is continuous and injective (and also the inverse is continous since it is the projection map). But then we know that $X_J$ is connected since every finite product is connected (if the components are connected).

Is this proof correct? The only thing i have to prove now is that $Y=\cup_{J\subset I,J\ finite}X_J$ is dense in X. How do I do that? Can someone help? Thank you

share|cite|improve this question

What you've done so far looks good. As you note correctly, it suffices to show that the set $Y=\bigcup_{J⊂I, J\text{ finite}}X_J$, which can also be described as $\{y\in X;\ y(i)=z(i)\textrm{ for almost all }i\in I\}$ is dense, since this means that $X=\overline Y$ and is thus connected, being the closure of the connected set $Y$. In order to do this, let $U=p^{-1}_{j_1}(U_{j_1})\cap\dots\cap p^{-1}_{j_n}(U_{j_n})$ be an open basis set. Now choose a $y$ such that $y(j)=p_j(y)\in U_j\ \forall j\in\{j_1,\dots,j_n\}$ and otherwise $y(i)=z(i)$. Then $y\in Y\cap U.$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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