Let X be a complete Alexandrov space with curvature ≥1 and radius 〉 π/2. We prove that any connected, complete, and locally convex subset without boundary in X also has the radius 〉 π/2.
基金Acknowledgements The authors would like to show their respect to the referees for their suggestions, especially on the form of the conclusion ＇rad（N） ≥rad（X） 〉 π/2＇ in Main Theorem （in the original version of the paper, the conclusion is ＇rad（N） 〉 π/2＇）. This work was supported in part by the National Natural Science Foundation of China （Grant Nos. 11001015, 11171025）.
文摘Let X be a complete Alexandrov space with curvature ≥1 and radius 〉 π/2. We prove that any connected, complete, and locally convex subset without boundary in X also has the radius 〉 π/2.
