Convergence in infinite(Tychonoff) product

Question

I have to prove the following

Proposition: Let $X=\prod X_i$ be a topological space and $\{(x_{n}^i)_{n\in \mathbb{N}} \}_{i\in I}$ a sequence in that space. $\{(x_{n}^i)_{n\in \mathbb{N}} \}_{i\in I}$ converges to $x$ if, and only if, $(x_{n}^i)_{n\in \mathbb{N}} $ converges to $x^i$ on $X_i$.

My attempt: $( \Rightarrow )$ Given a open nbhd of $x$, say $U$, the number of terms of the sequence which are not in $U$, say $X_n$ is finite, $n$. Also, considering the product topology, the i-th projection of $U$ , $\pi_i(U)=U_i $ is an open nbhd of the i-th projection of $x$, $\pi_i(x)=x_i$. As $\pi$ is a function, the number of elements of $\pi_i(X_n)$ is lesser or equal then n. As the open nbhd chosen was arbitrary we conclude that $(x_{n}^i)_{n\in \mathbb{N}}$ converges to $x_i$

($\Leftarrow$) Note that $ U=\Pi U_i=\Pi_{i\in I'}U_i \times\Pi_{i\notin I'}X_i$ where $i\in I $ and $I$ is finite, is an open nbhd of $x$ if for each $U_i, x_i\in U_i$ . As $(x^i_n)\rightarrow x_i$ the number of elements of the sequence which are not in $U_i$ is finite, say $n_i$. Then there exists a $n=\max\{n_i\}_{i \in I'} = |\{\{(x_{n}^i) \}_{i\in I} \notin U$}|. Therefore $\{(x_{n}^i)_{n\in \mathbb{N}} \}_{i\in I}$ converges to $x$ $\square$

Remarks: I'm not sure, specially in the $\Leftarrow$ either if I didn't miss arguments or if it's not right. I would like to make sure I'm quite precise.

Thanks in advance.

Answer 1

For the left to right direction simply use the theorem that continuous functions preserve limits and that $\pi_i$ is continuous.

Answer 2

($\Rightarrow$) Assume that there is an $j$ and the sequence $(x^{j}_n)$ does not converge to $x^j$. It means there is an open subset $U_j$ of $X_j$ such that $x^{j}_n\notin U_j$ for some proper $n$. Since $\pi_j$ is continious, then we have $\pi^{-1}_j(U_j)$ is open in $X$. We know $\{(x_{n}^i)_{n\in \mathbb{N}} \}_{i\in I}$ converges $x$. It means there exists $m\in\mathbb{N}$ such that $x_n\in \pi^{-1}_j(U_j)$ for all $m\leq n$. By the way $x_n=\{x^{i}_n\}_{i\in I}$ with your notations. It means $\pi_j(x_n)\in U_j$ so $\pi_j(x_n)=x^{j}_n\in U_j$ for all $m\leq n$. This is a contradiction with $(x^{j}_n)$ does not converge to $x^j$.

($\Leftarrow$) Take any open nbhd $U$ of $x=\{x^{i}\}_{i\in I}$. As you write above we can say $U=\Pi_{i\in I'}U_i \times\Pi_{i\notin I'}X_i=\bigcap_{i\in I'}\pi_i^{-1}(U_i)$ while $I'$ is finite. Since $(x^i_n)\rightarrow x_i$ then we have some $m_i\in \mathbb{N}$ where $i\in I'$ such that $x^i_n\in U_i$ for all $m_i\leq n$. Note that $\pi_i(x_n)=x^i_n$ and say $r=\max\{m_i\}_{i \in I'} $ then we have $x_n\in \pi^{-1}_i(U_i)$ while $r\leq n$ and $i \in I'$. So $x_n=\{x^{i}_n\}_{i\in I}\in \bigcap_{i\in I'}\pi^{-1}(U_i)=U$ while $r\leq n$. It means $\{(x_{n}^i)_{n\in \mathbb{N}} \}_{i\in I}$ converges to $x=\{x^{i}\}_{i\in I}$.