If $X$ is not compact, does this mean that Cone($X$) is not compact?

Question

I know that for $X$ a topological space, that if $X$ is compact then so is Cone($X$). Is the following identity also true: If $X$ is non-compact then Cone($X$) is not compact, if not can somebody give a counterexample.

Thanks in advance.

score 5 · Answer 1 · answered Dec 08 '13 at 23:48

5

The canonical map $X\to C(X)$ identifies $X$ with a closed subset of $C(X)$. Therefore, if $C(X)$ is compact, then $X$ is compact.

answered Dec 08 '13 at 23:48

Bruno Joyal

54,711

score 4 · Accepted Answer · answered Dec 08 '13 at 23:44

If $\{U_i\}$ is an open cover of $X$ with no finite subcover, then $\{Cone(U_i)\}$ is an open covering of $Cone(X)$. Since $U_i$ is homeomorphic to $\{(x,0)\mid (x,0)\in Cone(U_i))\}$ it follows that if $\{Cone(U_i)\}$ admitted a finite subcovering, say $Cone(U_1),\cdots, Cone(U_n)$, then we'd obtain a finite subcovering of the originl covering of $X$, a contradiction. Thus, if $X$ is not compact, then $Cone(X)$ is not compact.

If $X$ is not compact, does this mean that Cone($X$) is not compact?

2 Answers2