Exercise · ZW-CONNECTED-SPACE-008 · ★★☆ Intermediate
The open sets \(\mathbb{Q}\cap(-\infty,\sqrt2)\) and \(\mathbb{Q}\cap(\sqrt2,\infty)\) separate \(\mathbb{Q}\). Which of these pairs does not?
Choose one option.
A \(\mathbb{Q}\cap(-\infty,1)\) and \(\mathbb{Q}\cap(1,\infty)\) B \(\mathbb{Q}\cap(-\infty,\sqrt3)\) and \(\mathbb{Q}\cap(\sqrt3,\infty)\) C \(\mathbb{Q}\cap(-\infty,\pi)\) and \(\mathbb{Q}\cap(\pi,\infty)\) D \(\mathbb{Q}\cap(-\infty,-\sqrt2)\) and \(\mathbb{Q}\cap(-\sqrt2,\infty)\) Show answer A. \(\mathbb{Q}\cap(-\infty,1)\) and \(\mathbb{Q}\cap(1,\infty)\)← Back to the question
Explanation With a rational cut, the point \(1\in\mathbb{Q}\) lies in neither set , so they do not cover \(\mathbb{Q}\). With an irrational cut such as \(\sqrt3\), \(\pi\) or \(-\sqrt2\) they do cover, and both are open in \(\mathbb{Q}\).
A topological space \(X\) is connected if it cannot be written as the union of two disjoint nonempty open sets \(U\) and \(V\).
\[X=U\cup V,\ U\cap V=\varnothing\ \Rightarrow\ U=\varnothing\ \text{ or }\ V=\varnothing\]
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