# Basis of neighbourhoods of direct product $U \times U$

Let $$U \subseteq \mathbb{R} \subset \mathbb{C}$$ be open connected set in $$\mathbb{R}$$. Consider the basis of open complex neighbourhoods of $$U$$ and name it $$\mathscr{B} \left(U \right)$$ (i.e. for any neighbourhood $$W$$ of $$U$$ we can find a neighbourhood $$V \in \mathscr{B} \left(U \right)$$ that is contained in $$W$$). Now consider direct product $$U \times U \subseteq \mathbb{R}^2 \subset \mathbb{C}^2$$.

I would like to know, is there exist basis of neighbourhoods of $$U \times U$$, which contains only $$V \times V$$ for any $$V \in \mathscr{B} \left(U \right)$$ or in other words $$\mathscr{B} \left(U \times U \right) \stackrel{?}{=} \left\lbrace V \times V : V \in \mathscr{B} \left(U \right)\right\rbrace$$ is realy basis of neighbourhoods?

• Do you mean for $U$ to be open in the topology on $\mathbb{R}$ or $\mathbb{C}$? The only way for $U \subset \mathbb{R} \subset \mathbb{C}$ to be open and connected in the topology on $\mathbb{C}$ is for $U$ to be empty, assuming the standard embedding of $\mathbb{R}$ in $\mathbb{C}$. – Charles Hudgins Apr 25 at 18:05
• @Charles Hudgins, $U$ is open in $\mathbb{R}$, thanks! – Ann Apr 25 at 18:07
• @Ann What do mean by a neighborhood $W$ of a set $U$? Do you mean $W$ is a neighborhood of a point in $U$? – 0XLR Apr 25 at 20:00
• @ZeroXLR, If $S$ is a subset of topological space $X$ then a neighbourhood of $S$ is a set $V$ that includes an open set $U$ containing $S$. It follows that a set $V$ is a neighbourhood of $S$ if and only if it is a neighbourhood of all the points in $S$. Furthermore, it follows that $V$ is a neighbourhood of $S$ iff $S$ is a subset of the interior of $V$. The neighbourhood of a point is just a special case of this definition. – Ann Apr 25 at 20:14
• @Ann Well, if you require those neighborhoods $V$ to contain all points of $U$ and $\mathcal{B}(U)$ contained all such open neighborhoods of $U$ then it will work. Any neighborhood of the set $U \times U$ contains a standard basis element $V_1 \times V_2$ with $U \subseteq V_i$ where $V_i$ is open in $\mathbb{C}$. Then $V_1 \cap V_2$ is open in $\mathbb{C}$ and still contains $U$ and $U \times U \subseteq (V_1 \cap V_2) \times (V_1 \cap V_2)$. – 0XLR Apr 25 at 20:44

Not necessary even when $$U=\Bbb R$$. Indeed, consider a neighborhood $$X=\{(x, y,z,t)\in \Bbb C^2: (1+|x|)(1+|z|)(|y|+|t|)<1\}$$ of the set $$U\times U$$. Suppose to the contrary that there exists a neighborhood $$V$$ of $$U$$ such that $$V\times V\subset X$$. Pick an arbitrary point $$(x,y)\in V$$ with $$|y|>0$$. Then a point $$(1/y, 0)\in \Bbb R\subset V$$, so $$(x,y,1/y,0)\in V\times V\subset X$$. But $$(1+|x|)(1+1/|y|)|y|\ge 1+|y|>1,$$ a contradiction.