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Let $((E_i, \tau_i))_{i \in I}$ be a family of topological spaces, and let $\displaystyle E = \prod_{i \in I} E_i$ be its product space. Let $T$ be a subset of $E$, I wonder if we can conclude the following equivalence: $$x = (x_i)_{i \in I} \in \overline{T} \Longleftrightarrow x_i \in \overline{p_i (T)}, \forall i \in I.$$ Where $\overline{T}$ denotes the closure of $T$ and $p_i$ denotes the canonique projection from $E$ to $E_i$.

I've written the following arguement: $$\begin{array}{rl} x = (x_i)_{i \in I} \in \overline{T} & \Longleftrightarrow \forall U \in \mathscr{V}(x) \text{ with $U$ open in $E$, } U \cap T \ne \emptyset \\ & \Longleftrightarrow \forall \displaystyle \prod_{i \in I} U_i \in \mathscr{V}(x) \text{ with $U_i$ in $E_i$, } \Big(\prod_{i \in I} U_i\Big) \cap T \ne \emptyset\\ & \Longleftrightarrow \forall U_i \in \mathscr{V}(x_i) \text{ with $U_i$ open in $E_i$, } \forall i \in I, U_i \cap p_i(T) \ne \emptyset\\ & \Longleftrightarrow x_i \in \overline{p_i (T)}, \forall i \in I. \end{array}$$ Is that right?

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What you are trying to show is (very) false. The implication "$\Rightarrow $" is true and not hard to show.

But for the reverse implication, the set $$ T = \{(1,2), (2,1)\} \subset \Bbb {R} \times \Bbb {R} $$ provides a counterexample, since $(2,2)$ fulfils the second condition, but not the first.

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    Thanks! I know where the problem is: if $U_1 \cap p_1(T)$ and $U_2 \cap p_2(T)$ are not empty, we cannot conclude that $(U_1 \times U_2) \cap T$ is not empty! – 2017-02-27
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    Just now I have noticed that your counter-example is found in a space of finite dimension. But I've just observed that it is right in the space of infinite dimension. Do you think so? – 2017-03-07
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    @VictorChen: Almost the same counterexample works in infinite dimensions. Just consider $\mathbb{R}^2 \times \mathbb{R}^\mathbb{N}$ with the set $\tilde{T} = T \times \mathbb{R}^\mathbb{N}$ with $T$ as in the finite dimensional case. – 2017-03-09