Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/ib/paper-4/13g/i/solution
Past exam of the mathematics course of the University of Cambridge 2013 ib Paper 4 13G i Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-07
Fix , and let be the union of all connected subsets containing . Their union is connected: a separation of the union would place on one side, and every connected member meeting that side would have to lie entirely there. The other side would be empty. By the defining equivalence relation this union is exactly the connected component of .
The closure of a connected set is connected as well. Indeed, if its closure had a separation into relatively open nonempty pieces, density would make both pieces meet the original set and give a separation of that connected set. Therefore is a connected subset containing , so maximality gives . The reverse inclusion is automatic. Every connected component is connected and closed.
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