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Let $x_1, x_2, ... $ be a sequence of points of the product space $\prod X_{\alpha}$. Show that this sequence converges if and only if the sequence $\pi_{\alpha}(x_1)$, $\pi_{\alpha}(x_2)$, ... converges to $\pi_{\alpha}(x)$ for every $\alpha$

My Attempted Proof

Suppose $x_1, x_2$ converges to some point $x = (x_{\alpha})_{\alpha \in J} \in \prod X_{\alpha}$. Assume the sequence $\pi_{\alpha}(x_1)$, $\pi_{\alpha}(x_2)$ ... does not converge to $\pi_{\alpha}(x)$ for some $\alpha$. Put $\gamma = \alpha$ for this $\alpha$. Then we have $\pi_{\gamma}\left((x_{\alpha})_{\alpha \in J}\right) \neq x_{\gamma}$, reaching a contradiction.

Conversely suppose $\pi_{\alpha}(x_2)$, ... converges to $\pi_{\alpha}(x)$ for every $\alpha$. Let $\pi_{\alpha}\left(x_i\right) = x_{i_{\alpha}} $ denote the projection of $x_i \in \prod X_{\alpha}$ onto its $\alpha$th coordinate. Then we have $$x_{1_{\alpha}}, x_{2_{\alpha}}, .... \to x_{n_{\alpha}}$$ where $n > 0$. Hence we have $$(x_{i \in I})_{\alpha \in J} \to (x_{n})_{\alpha}$$ where $I$ is an arbitrary indexing set and $n = \sup I$. Thus we can see that $x_1, x_2, .. \to x_n = x$. $\square$

Is my proof correct? If so how rigorous is it? I was having some difficulty in proving this as it seemed something that intuitively obvious.

Any comments and criticism on the level of rigor of my proof is greatly appreciated.

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    Your notation is really bad, as you are using subscripts in the same location to mean two different things. Try using superscripts for one of them: The sequence is $\{x_n\}_{n=1}^{\infty}$, and denote $\pi_{\alpha}(y)$ by $y^{\alpha}$ for any $y$. So $x^\alpha_n :=\pi_\alpha(x_n)$. – 2017-01-20
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    There are plenty of topologies for which $\pi_\gamma(x_n)$ does not necessarity converge to $\pi_\gamma(x)$ just because ${x_n}$ converges to $x$. This convergence is a property of the product topology. The fact that your "proof" of this makes no mention of the properties of the product topology is a dead giveaway that it is flawed. the statement "then we have $\pi_\gamma((x_\alpha)_{\alpha \in J}) \neq x_\gamma$" is a non-sequitor. It does not follow from what is before. – 2017-01-20
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    Possible duplicate of [Convergence in product topology](https://math.stackexchange.com/questions/524481/convergence-in-product-topology) – 2018-09-12

1 Answers 1

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You do need the specifics of the product topology to show this, not just the continuity of projections.

So suppose $X = \prod_{\alpha \in A} X_\alpha$ has the product topology, and $x_n = (x^{(n)})_\alpha$ are points in $X$ for $n \in \mathbb{N}$. Then $x_n \rightarrow p = (p_\alpha)_\alpha \in X$ in the product topology iff for each $\alpha \in A$, $x^{(n)}_\alpha \rightarrow p_\alpha$ in $X_\alpha$.

The left to right implication of this just follows from continuity of $\pi_\alpha$, as $\pi_\alpha(x_n) = x^{(n)}_\alpha$ and $\pi_\alpha (p) = p_\alpha$ and continuity preserves convergence.

For the right to left implication, assume all components converge as stated. Then let $O$ be a basic open neighbourhood of $p$ in the product topology, so $O = \prod_\alpha O_\alpha$ where all $O_\alpha$ are open and $F = \{\alpha \in A: O_\alpha \neq X_\alpha \}$ is finite. Then for each $\alpha \in F$ ,pick $N_\alpha \in \mathbb{N}$ such that $$\forall n \ge N_\alpha: x^{(n)}_\alpha \in O_\alpha$$

by the definition of convergence in the compenent space $X_\alpha$ (as $O_\alpha$ is an open neighbourhood of the limit $p_\alpha$ there). Then as $F$ is finite (this is esesential here!) there is $N$ such that $N \ge N_\alpha$ for all $\alpha \in F$ (their maximum will do). Then for all $n \ge N$, we know that $x^{(n)}_\alpha \in O_\alpha$ for all $\alpha \in F$, which means that $x_n \in O$ for all $n \ge N$ as well (the only real conditions are on the coordinates from $F$). As we can find such $N$ for every basic open subset of the product, we have shown that $x_n \rightarrow p$, as required.

Note that this argument straightfowardly translates to nets as well, if you know about those. The way we found the $N$ from the $N_\alpha, \alpha \in F$ works in any directed index set not just $\mathbb{N}$.

A critique on your proof: you never use anything specific for the product topology , like their specific base elements, but this property fails for the box topology, which is another topology making the projections continuous (the only thing you're using if you think about it), or for many topologies on sequence spaces (essentially subsets of infinite products) in analysis as well.