Cones over noncompact spaces are not unions of paths I am trying to understand the following sentence
"When X is compact and Hausdorff (essentially, when X can be embedded in Euclidean space), then the cone CX can be visualized as the collection of lines joining every point of X to a single point. However, this picture fails when X is not compact or not Hausdorff, as generally the quotient topology on CX will be finer than the set of lines joining X to a point."
from the wikiepdia article about cones
. 
Can somebody give an explicit example of this phenomenon?
 A: Wikipedia is not always the most reliable source. The sentence you quoted is definitely not a precise statement.
1) "When $X$ is compact and Hausdorff (essentially, when $X$ can be embedded in Euclidean space)"
Compact Hausdorff spaces are in general not embeddable into an Euclidean space, thus "essentially" is totally misleading.
However, consider a subset $X \subset \mathbb{R}^n \times \{ 0 \} \subset \mathbb{R}^{n+1}$, pick a point $p \in \mathbb{R}^{n+1} \setminus \mathbb{R}^n \times \{ 0 \}$  and define
$$C_pX = \{ (1-t)x + tp \mid x \in X ,  t \in [0,1] \} = \bigcup_{x \in x} S(x,p)$$
where $S(x,p) =  \{ (1-t)x + tp \mid  t \in [0,1] \}$ is the line segment in $\mathbb{R}^{n+1}$ connecting $x$ and $p$. The space $C_pX$ is a "geometric cone" on $X$. It is a metrizable space, in particular it has a countable neighborhood base at $p$.
The map
$$h : CX \to C_pX, h([x,t]) = (1-t)x + tp$$
is always a continuous bijection. It is an easy exercise to show it is a homeomorphism provided $X$ is compact (here I used $CX = X \times [0,1]/X \times \{ 1 \}$) which is equivalent to the definition in Wikipedia).
2) "However, this picture fails when $X$ is not compact or not Hausdorff."
There is no picture unless $X \subset \mathbb{R}^n \times \{ 0 \}$. Such $X$ is automatically Hausdorff, but of course in general not compact.
We now show that the above map $h$ is no homeomorphism if $X$ is not compact - that seems to be what Wikipedia wants to say.
If $X$ is not compact, then we can find a sequence $(x_n)$ in $X$ which does not have an accumulation point in $X$. Since $p$ has a countable neighborhood base, we can find $t_n \in [0,1)$ such that $y_n = (1-t_n)x_n + t_np \to p$ in $C_pX \subset \mathbb{R}^{n+1}$. The set $A = \{ (x_n,t_n) \mid n  \in \mathbb{N} \}$ is closed in $X \times [0,1]$. We have $p^{-1}(p(A)) = A$, therefore $p(A)$ is closed in $CX$. But $h(p(A)) = \{ y_n \mid n \in \mathbb{N} \}$ is not closed in $C_pX$ (recall $y_n \to p$) which shows that $h$ is no homeomorphism.
Remark: The construction of "geometric cones" can be generalized a little.  For any $X \subsetneqq \mathbb{R}^m$ and any $p \in \mathbb{R}^m \setminus X$ we can define $C_pX$ as above. If $X \cap \{ (1-t)x + tp \mid x \in X ,  t \in (0,1] \} = \emptyset$, then $C_pX$ is a perfect geometric cone on $X$.
Such $p$ can always be found if the linear span of $X$ is a subspace of dimension $< m$. If the linear span is $\mathbb{R}^m$, then we cannot make a general statement about $C_pX$. Sometimes it is a geometric cone (for example when $X = S^{m-1}$ and $p = 0$), sometimes it is not (for example for $X = D^n$ and any $p$).
A: The general idea is this: if $X$ is a subspace of $\mathbb{R}^n$ then we can define a map from $X\times[0,1]$ to $\mathbb{R}^{n+1}$ by
$$
f(x,t) = \bigl<(1-t)\cdot x,t\bigr>
$$
For a fixed $x$ you get the line segment that connects $\langle x,0\rangle$ and $\langle\mathbf{0},1\rangle$.
This map is continuous and everywhere injective, except that it maps the top level $X\times\{1\}$ to one point.
This means that $f$ induces a map $\bar f:CX\to\mathbb{R}^{n+1}$; if $X$ is compact then $f$ is a closed map, hence a quotient map and so $\bar f$ is a homeomorphism.
If $X$ is not compact then this may not be the case, consider the case where $n=1$ and $X$ is the open interval $(0,1)$. The map $\bar f$ is not a homeomorphism as the set $U=\{(x,t):x\in X, t>x\}$ is open in $X\times[0,1]$ and it contains the top level. Therefore it determines a neighbourhood of the vertex in $CX$, but its image under $f$ does not determine a neighbourhood of $\langle0,1\rangle$ in the range of $f$.
You can draw all this: the range is a triangle, with base $X$ and top vertex $\langle 0,1\rangle$; the image of $U$ lies to the left of the parabola parametrised by $t\mapsto\langle t-t^2,t\rangle$.
