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Given a space $X$, the topological cone $CX$ is defined as the quotient $X\times I/(X\times \{1\})$.

If $X$ is a subset of $\mathbb R^n$, choose a point in $\mathbb R^{n+1}$ that does not lie in $\mathbb R^n$. The geometrical cone $aX$ is defined as $\{ta+(1-t)x|x\in X, t\in I\}$ equipped with subspace topology.

Proposition: If $X\subset \mathbb R^n$ is compact, then $CX$ is homeomorphic to $aX$.

Question: I wonder if it is possible to drop the assumption that $X$ is compact.

My teacher guessed that the conclusion remains to be true if $X$ is merely bounded, but didn’t provide any proof or counterexample.

fantasie
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  • You should take a look at this – Adam Chalumeau Jan 03 '20 at 07:59
  • Notice that the hypothesis of boundedness is irrelevant: in fact, a homeomorphism $\phi:\Bbb R^n\to\text{unit ball}$ induces a homeomorphism $CX\to C\phi[X]$ via the map $\pi\circ (\phi\rvert_X \times id)$ and a homeomorphism $aX\to a\phi[X]$ via the map $\phi\rvert_X\times id$. –  Jan 03 '20 at 08:09
  • No. If $X$ is non-compact, then $CX$ is not homeomorphic to $aX$. – Paul Frost Jan 03 '20 at 13:18
  • @PaulFrost Yes, thanks for your excellent answer. – fantasie Jan 04 '20 at 06:38

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