A sequence $\{x_n\}$ is convergent in $\prod X_\alpha$ iff the projection sequence $\{\pi_\alpha(x_n)\}$ converges in $X_\alpha$ for any $\alpha$

Question

Here is a proof I found.

To prove a sequence of points converges at a point $x$, I can show that for any neighborhood $U$ of $x$, there exists an integer $N$ such that for any $n \geq N$, $x_n \in U$. I don't understand why did he prove "for any subbasis set S for the product topology..."? This proof makes more sense to me.

Answer 1

He proved for any sub-basis set $S$ because that is enough. Let us see.
If $x \in V$, for some open set $V$, then $x \in U$ for some basic open set $U$ (because every open set is an union of basic open sets).
Now, if $x \in U$ for some basic open set $U$, then, since $U = U_1 \cap \cdots \cap U_k$ for some sub-basic sets $U_1,\ldots,U_k$, we have that $x \in U_i$, for $1 \leq i \leq k$.
If $x_n \in U_i$ whenever $n \geq N_i$ (what was proved), then $x_n \in U$ for $n \geq\max\{N_1, \ldots, N_k\}$.

(Overall the proofs using basis or sub-basis are typically shorter, but sometimes they don't give the same insight, and that was perhaps the reason for your confusion.)

Answer 2

Thanks for amrsa's answer. I have a better understanding now. The underline statement is actually false. A counterexample:

• $A=\{1,2\}, B=\{1,2\}, C= A \times B$. A subbasis for C is $\{ {(1,1), (1,2), (2,1)}, {(2,2)} \}$. Given a constant sequence $(1,2), (1,2), ....$ Since $\pi_1((1,2)) = 1$ and $\pi_2((1,2)) =2$, we have $\{\pi_n((1,2))\}$ converges to $n$ for $n=1,2$. Note that $\{(2,2)\}$ is also a subbasis, but there is no $(1,2)$ in $\{(2,2)\}$. So the existence part is not true for the subbasis element $\{(2,2)\}$.

I emailed him, and he told me that the underlined statement should be changed to

• For any subbasis set $S$ for the product topology on $\prod_{\alpha \in A} X_\alpha$ which contains the limit there exists $N\in \mathbb{N}$ such that $x_n \in S$ for all $n> N$.

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I think two proofs are essentially the same, but the proof in the picture is actually more complicated. I provide a detailed proof based on the proof on the picture. If I have any mistake, please pointing out. Thanks.

Proof: Suppose the sequence $\{\pi_\alpha(x_n)\}$ converges to $\pi_\alpha (x)$ for each $\alpha$. Let $\prod_{\alpha \in A} U_\alpha$ be an arbitrary neighborhood of $x$. Hence, $\prod_{\alpha \in A} U_\alpha$ is an element of the product topology of $\prod_{\alpha \in A} X_\alpha$. By definition of product topology, $$\prod_{\alpha \in A} U_\alpha = \text{"a finite intersection of subbasis elements"} = S_1 \cap S_2 \cap \cdots \cap S_k.$$ So, $x\in S_i$ for $1\leq i \leq k$. We need to prove that for any $i$, there exists $N\in \mathbb{N}$ such that $x_n \in S_i$ whenever $n\geq N$. By definition of subbasis, $S_i = \{\pi_\alpha^{-1}(U_\alpha):U_\alpha \text{ open in } X_\alpha\}.$ ...