Exercise 9, Section 17 of Munkres’ Topology [duplicate]

let $A\subseteq X$ and $B\subseteq Y$. Show that in the space $X\times Y$. $\overline{A\times B}=\overline{A} \times \overline{B}$.

My attempt: $\overline{A} \times \overline{B}$ is closed by exercise 3 of section 17. Since $A\times B \subseteq \overline{A} \times \overline{B}$,we have $\overline{A\times B}\subseteq \overline{A} \times \overline{B}$.

Conversely, let $x\times y \in \overline{A} \times \overline{B}$. $\mathcal{B}=\{P\times Q| P\in \mathcal{T}_X, Q\in \mathcal{T}_Y\}$ is basis for product topology $\mathcal{T}_{p}$. Let $\mathcal{N}_{x\times y}=\{ Z\in \mathcal{T}_p | x\times y \in Z\}$. So $x\times y\in Z= \bigcup_{i\in I}(U_i \times V_i)$; $(U_i \times V_i)\in \mathcal{B}, \forall i\in I$. So $x\times y\in U_j \times V_j$, for some $j\in I$. Since $x\times y \in \overline{A} \times \overline{B}$, we have $(U_j \times V_j)\cap (A\times B)= (U_j \cap A)\times (V_j \cap B)\neq \phi$. Since $(U_j \times V_j)\subseteq \bigcup_{i\in I}(U_i \times V_i)=Z$, we have $Z\cap (A\times B)\neq \phi$. Thus $x\times y\in \overline{A\times B}$. Is this proof correct?

A similar post exists $\operatorname{cl}(A \times B) = \operatorname{cl}_X(A) \times \operatorname{cl}_Y(B)$. I think my proof is little bit more explicit than the above link.

Solution 1:

In fact you can use that $\mathcal{B}_x$ is a neighbourhood base at $x$ in a space then $x \in \overline{A}$ iff $\forall B \in \mathcal B_x: B \cap A \neq \emptyset$ so in the product we only have to show that a basic set of the form $U \times V$ containing $(x,y)$ intersects $A \times B$, which is immediate when $x \in \overline{A}$ and $y \in \overline{B}$, as $(U \times V) \cap(A \times B)= (U \cap A) \times (V \cap B)$ is a product of non-empty sets. No need for the general $Z$.