Good evening everybody, I'm doing exercise II.4.2 in textbook Analysis I by Amann.
Could you please verify if my attempt contains logical mistakes/gaps! Thank you so much!
My attempt:
Lemma: The Cartesian product of sets is open (closed) iff each component set is open (closed).
$\Longrightarrow$: Assume $X_j$ is not connected. WLOG, assume $X_1$ is not connected. Then there are two nonempty disjoint open subsets $X^1_1, X^2_1$ of $X_1$ such that $X^1_1 \cup X^2_1 = X_1$. Then $X^1_1 \times \prod_{j=2}^{n} X_j$ and $X^2_1 \times \prod_{j=2}^{n} X_j$ are nonempty, open, and disjoint by Lemma. Moreover, $(X^1_1 \times \prod_{j=2}^{n} X_j )\cup (X^2_1 \times \prod_{j=2}^{n} X_j) = \prod_{j=1}^{n} X_j$. Then $\prod_{j=1}^{n}$ is not connected.
$\Longleftarrow$: Assume $\prod_{j=1}^{n}$ is not connected. Then there are subsets $X^1_j, X^2_j$ of $X_j$ for all $1 \le j \le n$ such that $\prod_{j=1}^{n} X^1_j$ and $\prod_{j=1}^{n} X^2_j$ are nonempty, disjoint, and open such that $(\prod_{j=1}^{n} X^1_j) \cup (\prod_{j=1}^{n} X^2_j) = \prod_{j=1}^{n} X_j$. By our Lemma, $X^1_k, X^2_k$ are nonempty, disjoint, and open such that $X^1_k \cup X^2_k = X_k$ for at least some $1 \le k \le n$. As such, $X_k$ is not connected.
Best Answer
Your proofs are fine, but really, you don't need the lemma. You can argue directly: it suffices to prove the claim for two connected spaces $X$ and $Y$. The result then follows by induction.
So choose any $(x_0,y_0)\in X\times Y$ and set $T_x=(X\times y_0)\bigcup (x\times Y).$ Then, $T_x$ is a union of connected sets with a point in common and so is connected. And since each $T_x$ has $(x_0,y_0)$ in common, $X\times Y=\bigcup _{x\in X}T_x$ is connected.