Let $X$ a topological space and $Y=\prod_{\alpha \in I}X_\alpha $ where $X_\alpha $ are copies of $X$. Define $$\pi_\alpha : Y\to X$$ the projection by $$\pi_\alpha (x_i )_{i\in I}=x_\alpha.$$ Is it always true that $$x_n\to x\iff \pi_\alpha (x_n)\to \pi_\alpha (x) \ \ ?$$
I proved that when $I$ is finite then yes. But what happen if $I$ infinite ?
It depend on your topology on $Y$. The implication always work since projection are continuous. But not the converse.
If $Y$ has the product topology, then it works (easy exercise)
If $Y$ has the box topology it doesn't work. Take for example $X=\mathbb R$ and $x_n=(1/n,1/n,1/n,...)$. Then $\pi_\alpha (x_n)\longrightarrow 0$ for all $\alpha $. Now, let $$U=\prod_{i=1}^\infty \left]-\frac{1}{i},\frac{1}{i}\right[,$$ since $\left]-\frac{1}{i},\frac{1}{i}\right[$ is open for all $i$, then $U$ is open in $Y$. But there is no $N$ s.t. $x_n\in U$ for all $n\geq N$. Indeed, if such $N$ exist, then $\frac{1}{N}=\pi_{i}(x_N)<\frac{1}{i}$ for all $i$ and thus $\frac{1}{N}=0$ which is impossible.