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Is the set of points in the plane whose coordinates are either both irrational, or both rational connected?

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    Its ok,perhaps somebody else might be aware of it!2019-01-30

1 Answers 1

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Let $S$ be your set, and suppose $S = U \cup V$ where $U$ and $V$ are disjoint, and are both closed and open in $S$. Note that if $x$ and $y$ are both rational, then the diagonal lines $\{(x+t, y+t): t \in {\mathbb R}\}$ and $\{(x+t,y-t): t \in {\mathbb R}\}$ are subsets of $S$. Using a path of diagonal line segments, it is possible to get from any point of ${\mathbb Q} \times {\mathbb Q}$ to any other while staying in $S$. Therefore one of $U$ and $V$, let's say $U$, contains all of ${\mathbb Q} \times {\mathbb Q}$. But ${\mathbb Q} \times {\mathbb Q}$ is dense in $S$, and $U$ is closed in $S$ so $U = S$.

For extra credit, show that $S$ is path-connected. In fact, if a < b and c < d with $(a,c)$ and $(c,d)$ in $S$, there is a continuous increasing function $f: [a,b] \to [c,d]$ such that $f(x)$ is rational if and only if $x$ is rational.

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    Take two-sided increasing sequences of rationals $x_n$ and $y_n$ with $x_n \to a$ and $y_n \to b$ as $n \to -\infty$, $x_n \to c$ and $y_n \to d$ as $n \to +\infty$, and join $(x_n, y_n)$ to $(x_{n+1},y_{n+1})$ with straight line segments.2018-01-08