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It is well known that if $(X,\tau_X)$ and $(Y,\tau_Y)$ be two connected topological spaces then $(X\times Y,\tau)$ is also connected where $\tau$ is the product topology on $X\times Y$.

However, all the proofs that I have seen either uses an argument like this or this or this.

But after seeing Theorem 26.15 from here I was wondering whether it is possible to prove the following using the simple chain condition.

Definition of simple chain connecting two points. A simple chain connecting two points $a$ and $b$ of a space $X$ is a sequence $U_1,\ldots,U_n$ of open sets such that $a\in U_1$ only and $b\in U_n$ only and $U_i\cap U_j\ne\emptyset$ iff $|i-j|\le 1$.

More specifically,

  1. Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be two connected topological spaces. Show that $(X\times Y,\tau)$ is also connected by proving that given any two points $(x_1,y_1),(x_2,y_2)\in X\times Y$ and any open cover $\mathscr{U}$ of $X\times Y$ there exists a simple chain from $\mathscr{U}$ connecting $(x_1,y_1)$ and $(x_2,y_2)$.

  2. Let $X_i$, $i \in I$ is a family of nonempty topological spaces, all of which are connected spaces. Show the product space $\displaystyle\prod_{i \in I} X_i$, endowed with the product topology, is also connected by proving that given any two points ${\bf{x},\bf{y}}\in \displaystyle\prod_{i \in I} X_i$ and any open cover $\mathscr{U}$ of $\displaystyle\prod_{i \in I} X_i$ there exists a simple chain from $\mathscr{U}$ connecting $\bf{x}$ and $\bf{y}$.

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Your approach doesn't work. It is not true that a topological space $X$ is connected if and only if, for any two points $x,y\in X$, there is a simple chain connecting them. Because for any two points $x,y\in X$ (whether $X$ is connected or not) you can always find such a chain: just take $n=1$ and $U_1=X$.

José Carlos Santos
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  • The definition states $x \in U_1$ only and $y \in U_n$ only. Presumably, this means $x \not \in U_i$ for $i \neq 1$, and similarly for $y$. – lisyarus Mar 31 '18 at 09:47
  • @lisyarus I don't understand your objection. It is obvious that $x\in U_1$ only and $y\in U_1$ only. There is no $U_i$ for $i>1$. – José Carlos Santos Mar 31 '18 at 09:50
  • You are right. I mean, probably it is meant that $n \geq 2$? Still, your answer is correct. – lisyarus Mar 31 '18 at 09:56
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    Looking in the book, the theorem is actually that $X$ is connected if and only if, for any open cover $\mathscr{U}$, there is a simple chain connecting any two elements whose elements belong to $\mathscr{U}$. – Alex Zorn Mar 31 '18 at 09:57
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    @AlexZorn Right! It is essential to add this part concerning the open covers. – José Carlos Santos Mar 31 '18 at 09:58
  • Sorry, but I don't understand how your post answers my questions. What am I missing? –  Mar 31 '18 at 13:26
  • @user170039 I proved thattevery topological space (connected or not) satisfies the condition that you mentioned. Therefore, the condition cannot be used to prove that a space is connected. – José Carlos Santos Mar 31 '18 at 13:30
  • @JoséCarlosSantos: I have made a small edit to make my question more precise. Please take a look at it. –  Mar 31 '18 at 13:30
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    @user170039 THat is not a small edit; it's a huge one. Anyway, my answer applied to your question as it was stated at the time I've provided it. – José Carlos Santos Mar 31 '18 at 13:34