Can someone explain me this proof please:

We have $\prod_{i\in I} \overline{A_i}=\cap_{i\in I} F_i$ where $F_i=\prod_{j\in I} B_j$ with $B_i=\overline{A_i}$ and $B_j=E_j$ if $j\neq i$. We have that $E\setminus F_i=\prod_{j\in I} U_j$ with $U_i=E_i\setminus \overline{A_i}$ and $U_j=E_j$ if $j\neq i$. Then $E\setminus F_i$ is open in $E$. So, $F_i$ is closed in $E$, from where $\prod_{i\in I} \overline{A_i}$ is closed in $E$ which contains $A$. Then we have $\overline{A}\subset \prod_{i\in I} \overline{A_i}$.

In the other hand, let $x=(x_i)_{i\in I}\in \prod_{i\in I} \overline{A_i}$ and $U=\prod_{i\in I} U_i$ an elementary open of $E$ containing $x$. For each $i\in I,$ $U_i$ in an open of $E_i$ and there exists a finite subset $J$ of $I$ such that for all $i\in I\setminus J$, we have $U_i=E_i$. For all $i\in I$, we have $x_i\in \overline{A_i}\cap U_i,$ then there exist $a_i\in A_i\cap U_i$, where $a=(a_i)_{i\in I}\in A\cap U$. then we have $A\cap U\neq\emptyset.$ And then we have,$x\in \overline{A}$. Finally $\overline{A}=\prod_{i\in I}\overline{A_i}.$

Thank you.


I assume $A = \prod_i A_i$ (this is not stated), where all $A_i$ are non-empty.

We define $F_i$ as $(\pi_i)^{-1}[\overline{A_i}]$, which is closed as the inverse image of a closed set under a continuous projection.

$(x_i)$ being in $\prod_i \overline{A_i}$ means that for all $i$, $x_i \in \overline{A_i}$, so $x \in (\pi_i)^{-1}[A_i] = F_i$ so that $\prod_i \overline{A_i}$ is closed as the intersection of the closed sets $F_i$.

As $A$ is a subset of this closed set, its closure (the smallest closed set that contains $A$) is a subset of it as well.

If $x \in \prod_i \overline{A_i}$, let $U = \prod_i U_i$ be a basic open set around $x$, wich depends on finitely many coordinates $J$. For each $j \in J$ we know that $x_j \in \overline{A_j}$ and $x_j \in U_j$, so we have $a_j \in A_j \cap U_j$ for all $j \in J$. Pick any $a_i \in A_i$ for $i \notin J$. Then the $(a_i)$ is in $U$ and in $A$, so every neighbourhood of $x$ intersects $A$, so $x \in \overline{A}$. this shows the other inclusion.

  • $\begingroup$ Please is the pruduct can be infinit please $\endgroup$ – Vrouvrou Nov 15 '15 at 6:49
  • $\begingroup$ Yes, the proof works for all products $\endgroup$ – Henno Brandsma Nov 15 '15 at 7:12
  • $\begingroup$ But if i prove it by two inclusions for $A=A_1\times A_2$ how i can generelize it ? (Ps we have study only finite product) $\endgroup$ – Vrouvrou Nov 15 '15 at 7:21
  • $\begingroup$ @Vrouvrou the exact same proof works for finite sets $I$ as well. Try it. I use nothing at all about $I$. The only simplifying thing is that $U$ can directly be written as $U_1 \times U_2 \ldots \times U_n$ instead, if we use $\{1,\ldots,n\}$ as the index set. $\endgroup$ – Henno Brandsma Nov 15 '15 at 10:50
  • $\begingroup$ If i prove like this: $(x,y)\in \overline{A_1\times A_2}\Longleftrightarrow\forall V_X\in \mathcal{V}_X, V \cap(A_1\times A_2)\neq \emptyset\\\Longleftrightarrow \forall V_1\in \mathcal{V}_x, \forall V_2\in \mathcal{V}_y, (V_1\times V_2)\cap (A_1\times A_2)\neq \emptyset$ $\endgroup$ – Vrouvrou Nov 15 '15 at 20:20

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