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I've been trying to tackle this problem for some while now, but don't know how to start correctly. I know that the cone on $(0,1)$ is given by $$\text{Cone}((0,1)) = (0,1) \times [0,1]/((0,1)\times\{1\}).$$ But how do I show that it can not be embedded in an Euclidean space? Cause for me it looks like it is possible.(Open cylinder with the "ceiling" collapsed to one point. I'm guessing that the problem for me also lies in what a quotient really is, cause I can't really get a good feeling for it.

I don't want the answer, I just want a push in the right direction so that I can think about how to solve it.

Edit: New insight, when thinking about the cone, it should be something like this (I guess) but this would mean that it can be embedded in $\mathbb{R}^2$ I think, which contradicts the question.Cone as I think it should be

Thanks

user112167
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  • Is embeddness means a map $C \to \mathbb R^n$ such that the topology of $C$ agrees with the subspace topology of the image in $\mathbb R^n$? –  Dec 02 '13 at 23:38
  • If $X\subset \mathbb R^k$ then $C(X)$ embeds in $\mathbb R^{k+1}$. So, yes it embeds. However, it is not homeomorphic to an open hemisphere. The cone point sits on the boundary. – Cheerful Parsnip Dec 02 '13 at 23:41
  • The strange thing is, we haven't discussed embedness yet, and we have to look intuivitly to it (By an exact model or such). But you ay it embeds, but we have to show that it cannot be embedded, how is this possible? – user112167 Dec 02 '13 at 23:43
  • I agree that the cone does embed. It might help if you could share some more of the context in which the problem was set. – Kevin Carlson Dec 03 '13 at 00:04
  • I don't see why its not homeomorphic to the hemisphere. How is boundary defined in this context? Boundary is something either defined for manifolds, or for subsets of topological spaces - not topological spaces. – Stephen Montgomery-Smith Dec 03 '13 at 00:06
  • I have the feeling that $C$ cannot be embedded. If it is embedded, it seems that every open set containing the vertex must contain $(0,1) \times [0,\epsilon]/\sim$ for some $\epsilon$, but there are more open set in $C$ (e.g. $(.5,1)\times [0,1]/\sim$ –  Dec 03 '13 at 00:16
  • There isn't really a setting around it. It's kind of an extra exercise: when you finished the other homework, try this some more difficult exercise. I also tried to think somemore about it, and think it should be a kind of triangle with the bottom side open. Cause the interval $(0,1)$ is open. – user112167 Dec 03 '13 at 00:49
  • @John $(0.5,1)\times[0,1]/\sim$ isn't open in the quotient topology, because its inverse image includes all of $(0,1)\times{1}$. – Andreas Blass Dec 03 '13 at 01:32
  • This cone does embed in $\mathbb R^2$, just as shown in the picture. Its image consists of an open triangular region, plus one side (without its endpoints), plus the vertex opposite that side. – Andreas Blass Dec 03 '13 at 01:35
  • @AndreasBlass: You are right. Thanks for that. –  Dec 03 '13 at 01:41
  • So the question is wrong? Cause the triangle I just drawn is what I make of it, following the definition of the cone. – user112167 Dec 03 '13 at 01:45
  • @AndreasBlass: I don't understand. Why is the quotient even first countable? – Niels J. Diepeveen Dec 04 '13 at 00:37
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    @NielsDiepeveen I think you're right. The non-compactness of $(0,1)$ prevents the "obvious" (but false) argument for first-countability at the vertex of the cone. I therefore retract my claim that the cone embeds in the plane (or in any metric space). – Andreas Blass Dec 04 '13 at 23:59
  • Could you please explain that in some detail, cause I still do not get it. – user112167 Dec 05 '13 at 00:03
  • I retract my claim as well. I agree that it can't be embedded. – Cheerful Parsnip Dec 05 '13 at 14:35

3 Answers3

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Euclidean spaces are metric spaces, so in order for the cone to be embeddable in one of them, it must be a metrizable space. As far as I can make out, the cone of a space is metrizable if and only if the space itself is metrizable and compact. Obviously $(0, 1)$ is metrizable but not compact.

I have been looking for a reference for this equivalence, but could not find one. The best support for the necessity of compactness that I did find is exercise 23K in Willard's General topology, which says (among other things)

Let $f$ be a closed continuous map of a metric space $M$ onto a space $Y$.

...

The following are equivalent

  • $Y$ is metrizable
  • $Y$ is first countable
  • For each $p \in Y$, $f^{-1}(p)$ has compact frontier
Cheerful Parsnip
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Niels J. Diepeveen
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The cone $C(J)$ where $J=\mathrm{int}(I)$ is not first countable. Consider the subspace $$B:=S\times I:=\left\{\frac1n\middle|n\in\Bbb N\right\}\times I$$ of $J\times I$. It is closed, and each closed and saturated subset $A$ of $B$ is either disjoint from $J\times\{1\}$, in this case it is saturated in $J\times I$, or it contains $J\times\{1\}\cap B$, but in that case its saturation is $A\cup J\times\{1\}$ which is closed. So each closed and saturated subset of $B$ is the intersection of a closed and saturated set in $J\times I$ with $B$. Therefore the restriction of the quotient map $q:J\times I\to C(J)$ to $B$ is a quotient map and $C(S)$ is a subspace of $C(J)$.

Now, $C(S)$ is a CW complex with infinitely many cells meeting the apex of the cone, so it is not first countable. Hence the superspace $C(J)$ cannot be a subspace of a metric space.

Stefan Hamcke
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0

Let me add a little bit of context and change the terminology such that ‘automatic thinking’ gets it right.

Let $\mathrm{C}(X)$ denote the (real, geometric, true...) cone on $X$. Its points are the pairs $(\lambda, x) \in \mathbf R_+^* \times X$ and a special point $0$.

An open $U$ of the real cone is either an open of $\mathbf R_+^* \times X$ if it does not contain $0$ and if it does contains $0$, it is additionally required to contain an open of the form $(0, \varepsilon) \times X$.

Can you show that if $X$ is embeddable in $\mathbf R^n$, then the real cone $\mathrm{C}(X)$ is embeddable in $\mathbf R^{n+1}$?

Let $\mathrm{CR}(X) = \mathbf R_+ \times X/0 \times X$ be the collapsed rectangle. Now, when $X$ is compact and separated, it happens to be true that $$ \mathrm{C}(X) \simeq \mathrm{CR}(X) $$ but not in general. Can you show that the two are different in the case $X = (0,1)$?

Damien L
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