Let $$\begin{align*}X=&\left\{(x,y,z)\in \mathbb{R}^3 \mid x^2+y^2=1, 0\leq z \leq 1 \right\}\\&\cup \left\{ (x,y,z)\in \mathbb{R}^3 \mid x^2+y^2\leq 1, z\in \{0,1\} \right\}\end{align*}$$ equipped with the induced topology by the usual topology in $\mathbb{R}^3$. Consider the quotient space obtained by identifying all points of the base of the cylinder $B=\left\{ x^2+y^2\leq 1 \mid z=0 \right\}$. Let $p$ stand for the canonical projection from $X$ to $X/B$. Study the compactness of both $X$ and $X/B$.

I think I am missing something here but I don't know what.

First, the space $X$ is compact. It is closed and bounded or it can be written as the product $\mathbb{S}^1 \times [0,1]$ union the closed discs $x^2+y^2\leq 1,\ z=0,1$. But then $p$ is a continuous surjection, so $X/B$ is also compact (in the quotient topology).

I've had always trouble when quotient topology questions appear, so I wouldn't be surprised if I missed something. Thanks in advance!

  • $\begingroup$ @BrianM.Scott Ups, I mistranslated it. It should say "identify all points of the base of the cylinder $B=\left\{ (x,y,z) \mid x^2+y^2\leq 1, z=0 \right\}$". So instead the quotient $X/B$ should be a cone. $\endgroup$ – user313212 Sep 10 '16 at 18:41
  • $\begingroup$ Thanks: I thought that that might be the case. (Out of curiosity, translated from what? You had proyection originally, so I’m guessing that it might have been Spanish.) $\endgroup$ – Brian M. Scott Sep 10 '16 at 18:45
  • $\begingroup$ Your argument is fine if you’re using what I consider the correct definition of compact, which does not require the space to be Hausdorff. If your definition of compactness includes a requirement that the space be Hausdorff, then you’ll also need to show that $X/B$ is Hausdorff, which in this case is very easy. $\endgroup$ – Brian M. Scott Sep 10 '16 at 18:53
  • $\begingroup$ @BrianM.Scott Thank you! And exactly, it was Spanish! The j/y is something I need to be careful about. At least I wrote surjection correctly! $\endgroup$ – user313212 Sep 10 '16 at 19:03
  • $\begingroup$ For future reference, remember that the quotient topology on $X/B$ is defined as the strongest topology on the set $X/B$ such that the quotient map $X\to X/B$ is continuous. So the open sets of $X/B$ are precisely the images, under the quotient map, of the open sets of $X.$ $\endgroup$ – DanielWainfleet Sep 13 '16 at 7:26

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