Prove that the cartesian product of topological manifolds is again a topological manifold. I have the following problem:

Let $I$ be a finite index set. Show that for top. manifold $M_i$ their product $M=\prod_{i\in I} M_i$ is again a top. manifold.

Proof
Let $(M_i,T_{M_i})_{i\in I}$ be topological manifolds. Let us consider $M=\prod_{i\in I} M_i$ with its product topology, i.e. $(M,T)$. I need to show the following three points:

*

*$\forall p\in M$ there exists $U\in T$ such that $p\in U$ and $x:\mathbb{R}^m\rightarrow U$ is a homeomorphism. Let $p=(p_1,...p_{dim(I)})\in M$. Since $(M_i,T_{M_i})_{i\in I}$ are top. manifolds we know that all $p_k\in O_k\in T_{M_k}$ for all k. Then let us define $U=\prod_{i\in I} O_i$ then $p$ is clearly in U and $U\in T$. Furthermore we know about the existance of local charts $x_k:\mathbb{R}^m\rightarrow O_k$. Therefore let us define $x:\mathbb{R}^{dim(I)} \rightarrow U$ such that $x(q)=\left(x_1\left(q_1\right),...,x_{dim\left(I\right)}\left(q_{dim\left(I\right)}\right)\right)$. But since all $x_k$ are homeomorphism we know that also $x$ is a homeomorphism.


*$T$ on M is second countable. Let us remark that all the topologies $T_{M_i}$ are second countable. I denote the basis of $T_{M_i}$ as $B_{M_i}$. Then the basis $B$ of $T$ is the union of the $B_{M_i}$, which is finite since finte intersection of finite elements are finite.


*The topology $T$ on $M$ is Hausdorff. Let $p,q\in M$ such that $p\neq q$. Then $p_k\neq q_k$ for all k. Since $M_k$ are Hausdorff we can finde disjoint $O_k,Q_k$ such that $p_k\in O_k$ and $q_k\in Q_k$. Take $O=\prod_{k\in I} O_k$ and $Q=\prod_{k\in I} Q_k$ then $p\in O$, $q\in Q$ and $O\cap Q=\emptyset$
I'm not sure if this works and I would appreciate it if someone could take some time to discuss it.
Thank you a lot!!
 A: Your approach is correct, but you

*

*use terminology which is undefined


*some of your argument are somewhat imprecise.



*

*$\dim(I)$ is undefined. It certainly denotes the number of elements of $I$. You should take $I = \{1,\ldots,r\}$ and use $r$ instead of $\dim(I)$. Note that this is not the most general form a finite set, but you write $p =(p_1,\ldots, p_r)$ which shows that you really work with the above special case. And, by the way, $\mathbb{R}^{\dim(I)}$ does not have anything to do with that what you need.


*You should say that $T_{M_i}$ denotes the topology on $M_i$.


*It seems that you understand of a (local) chart on $M_i$ as a homeomorphism $\mathbb R^{m_i} \to U$, where $U \subset M_i$ is open. Note that $m_i$ is the dimension of $M_i$, you cannot work with a common $m$ as you do. Your concept of chart is somewhat unusual, the standard interpretation is that of a homeomorphism $U \to V$, where $U \subset M_i$ and $V \subset \mathbb R^{m_i}$ are open. Charts in your sense are sometimes called local coordinate systems (though in general one allows arbitrary open $V \subset \mathbb R^{m_i}$ as domains). Anyway, all these concepts are equivalent - but it deserves an explanation what your interpretation is.


*"Since $(M_i,T_{M_i})_{i\in I}$ are top. manifolds we know that all $p_k\in O_k\in T_{M_k}$ for all k. Then let us define $U=\prod_{i\in I} O_i$ then $p$ is clearly in U and $U\in T$. Furthermore we know about the existance of local charts $x_k:\mathbb{R}^m\rightarrow O_k$. ..."
In the first sentence you not tell us what the $O_i$ are. You explain it in the last sentence, and this is an illogical order. You should argue as follows:
We know about the existence of local charts $x_k:\mathbb{R}^{m_i}\rightarrow O_k$ with $p_k \in O_k$. Let us define $U=\prod_{i\in I} O_i$. Then $p$ is clearly in U and $U\in T$. Let us define $x:\mathbb{R}^{n} \rightarrow U$ such that $x(q)=\left(x_1\left(q_1\right),...,x_r(q_r)\right)$. Here $n = \sum_{k=1}^r m_k$. But since all $x_k$ are homeomorphism we know that also $x$ is a homeomorphism.


*"I denote the basis of $T_{M_i}$ as $B_{M_i}$. Then the basis $B$ of $T$ is the union of the $B_{M_i}$, which is finite since finte intersection of finite elements are finite." Correct it to
Let $B_{M_i}$ be a countable basis of $T_{M_i}$. Then a basis $B$ of $T$ is given as the set of all products $\prod_i U_i$ with $U_i \in B_{M_i}$.


*"The topology $T$ on $M$ is Hausdorff. Let $p,q\in M$ such that $p\neq q$. Then $p_k\neq q_k$ for all k. Since $M_k$ are Hausdorff we can finde disjoint $O_k,Q_k$ such that $p_k\in O_k$ and $q_k\in Q_k$. Take $O=\prod_{k\in I} O_k$ and $Q=\prod_{k\in I} Q_k$ then $p\in O$, $q\in Q$ and $O\cap Q=\emptyset$." Correct it to
Then $p_k\neq q_k$ for some $k$. Since $M_k$ are Hausdorff we can finde disjoint $O_k,Q_k$ such that $p_k\in O_k$ and $q_k\in Q_k$. Take $O=\prod_{k\in I} O_k$ and $Q=\prod_{k\in I} Q_k$, where the other $O_i,Q_i$ can be chosen as arbitrary open neighborhoods of $p_i, q_i$. Then $p\in O$, $q\in Q$ and $O\cap Q=\emptyset$."
