Correct answer: (C) If both sets are closed, their sum is closed.
Explanation
C fails. In ℝ² let $E_1=\{(x,1/x):x>0\}$ and E₂ = ℝ×{0}. Both are closed, but their sum is ℝ×(0,∞), not closed. A follows by taking a union of translates of the open set; B by combining convex combinations; D is the product theorem for connected spaces. **C**.
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