Homogeneous space minus a point If $X$ is homogeneous and $p\in X$, then is $X\setminus \{p\}$ necessarily homogeneous?
This seems to work with all the simple examples I've tried. I would be interested in any counterexamples.  Or if there is a theorem here, you can assume we are dealing with metric spaces.
EDIT: Greg's pointed out that $X=2\times S^1$ is an example of a homogeneous space such that $X\setminus \{p\}$ is not homogeneous, for any $p$. So I will modify my question.
If $X$ is homogeneous and connected, and $p\in X$, then is $X\setminus \{p\}$ necessarily homogeneous?
 A: A non-Hausdorff example is the real line with the Poset topology, where the non-trivial open sets are $(-\infty,a)$ for $a\in\mathbb R$. 
The "Long Line" will give a a Hausdorff example.
Let $\omega_1$ be the first uncountable ordinal.
Define $L=\omega_1\times [0,1)$, and define an order $\leq$ as $(w_1,x_1)<(w_2,x_2)$ of $w_1<w_2$ or ($w_1=w_2$ and $x_1\leq x_2$.) Define the topology as the order topology on this set.
Remove the left point $(0,0)$. We might call this resulting space, $L_1$, the "open long line."
This space is homogeneous, because given any $x_1,x_2\in L_1$, find a $y>x_1,x_2$. The space $(0,y]$ can be seen to be homeomorphic to $(0,1]$ (this is where you need $\omega_1$ to be minimal.) This means that any $x_1,x_2<y$ can be sent to each other by some homeomorphism $L\to L$ which fixes all $y'\geq y$.
But if you remove $y$ from $L_1$, then $(0,y)$ is just an open real interval, and $(y,\infty)$ is homeomorphic to the original $L_1$. So the points in $L_1\setminus\{y\}$ have two different classes.
Both of these have the property that homeomorphisms must preserve an order - if $x<y$ then $f(x)<f(y)$.
