Is $\Bbb Q\times(\Bbb R\setminus\Bbb Q)$ connected?

Question

Reading the definition for a set to be connected, I found that my intuition breaks down and I wonder whether the following set is connected?$ \def\Q{\Bbb Q} \def\R{\Bbb R} \def\c{^{\mathsf C}} \def\-{\!\setminus\!} \def\X#1#2{\mathop{\LARGE \times}_{#1}^{#2}} \def\ue{\mathrm{u.\!e.}} $ The complement is denoted as $\cdot\c$

Let $M\subset\R$ be countable and dense in $\R$. Is $X=M\times M\c$ connected?

I'd guess that the answer is independent of $M$, i.e. it does not matter whether it is the Rational numbers, the Algebraic numbers, or some real number-field etc.?

In the case that it's not connected:

• Can connectedness be achieved by adding more copies of $M\c$? Like is $X_n(\Q)$ connected for $$ X_n(M) = M\times\X 1 n M\c $$

• Can connectedness be achieved by using a set $M$ such that $M$ and $M\c$ are "uncountable everywhere"? Let "$M$ is u.e. in $\R^n$" be defined as:$$ M\subset\R^n \text{ is }\ue \quad\iff\quad (S\subset\R^n, S \text{ open } \implies |S\cap M| > \aleph_0)$$

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Note: There is an other question if $\Q^n \cup (\R\-\Q)^n$ is connected, but I think that does not help here because it's union, not Cartesian product?

Answer 1

$X\times Y$ is connected if and only if both $X$ and $Y$ are connected. The "only if" part is a direct consequence of projection maps $\pi_X(x,y):=x$ and $\pi_Y(x,y):=y$ being continuous (and of the empty topological space not being connected). The "if" part requires a bit more work, but it's in all books. For the same reason, given a family of topological spaces $\{(X_i,\tau_i)\}_{i\in I}$, their product $\prod_{i\in I}X_i$ is connected if and only if all of them are. Therefore:

• the subsets $M\subseteq\Bbb R$ such that $M\times (\Bbb R\setminus M)$ is connected are exactly the intervals in the form $[a,\infty)$, $(a,\infty)$, $(-\infty,a]$ or $(-\infty,a)$ for some $a\in\Bbb R$ (otherwise $\Bbb R\setminus M$ either is empty or it has more than one connected component)

• "adding" more copies of a factor to a disconnected product never improves the situation.

Answer 2

Suppose that $p,q \in \mathbb{R}$ are such that $p \notin M$ and $q \notin M^c$. Such numbers clearly exist. Then, $(M \times M^c) \cap (\{p\} \times \mathbb{R}) = (M \times M^c) \cap (\mathbb{R} \times \{q\}) = \emptyset$. Hence you can write $M \times M^c$ as the product of four (and hence also two) disjoint opens.

On your first bullet point: if a set is connected, then all of its projections must be too. The answer to your second bullet point is also no, because I can just take the product of the nonalgebraic numbers by themselves. That's also not connected by the argument above.

Answer 3

Let $a$ be any point not in $M$. Then $\{(x,y): x<a\}$ and $\{(x,y): x>a\}$ are disjoint open sets whose union covers $M \times M^{c}$. Hence $M \times M^{c}$ is not connected.

The same argument works for $M \times M^{c} \times M^{c}\times ... \times M^{c}$.