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For a compact Riemannian manifold, Since the curvature tensor is continuous, we know that the sectional curvature is bounded, i.e. bounded above and below. Now let $M$ be a compact Alexandrov space with curvature bounded below, then is the curvature "bounded above"?

Since I don't quite understand the definition of curvature bounded above. I just try to construct a counter example: Beginning with a sphere $S^2$, we cut off countable small holes, and glue them with cones which are sharper and sharper (i.e. the curvature goes to $+\infty$), finally we get $N$.

However $N$ is not an Alexandrov space with curvature bounded below, since at the boundary of the holes, we can check that $N$ doesn't satisfy the definition i.e. the sum of the three comparison angle is less than $2\pi$.

So does someone have a counter example? Or can we smooth the neighborhoods of the holes, to make $N$ has a lower curvature bound, for exmaple $-1$?

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    $\begingroup$ First read the definitions :) Then take the surface of convex polytope; it has curvature bounded below, but no upper curvature bound. $\endgroup$ – Anton Petrunin Aug 28 '15 at 10:30