Show that the cone of the open interval (0, 1) can not be embedded in any Euclidean space 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.
Thanks
 A: 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
  

A: 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.
A: 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 $X = (0,1)$?
