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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?

Forever Mozart
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    If you allow non-connected homogeneous spaces, the answer seems to be no: let $X$ be the disjoint union of two circles, for example. – Greg Martin Feb 05 '16 at 09:41
  • Ok, that is good to know. I was thinking of connected spaces, so I will change my question. – Forever Mozart Feb 05 '16 at 22:28
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    The "long line" is a good example of a space that is homogenous but where removing a point leaves something non-homogeneous. Basically, if you remove a point from the long line, then it splits it into disconnected components, one of which is a standard interval, and the other is homemorphic to the "long line." – Thomas Andrews Feb 06 '16 at 02:28
  • yes that works. so it's not even true for connected spaces. – Forever Mozart Feb 06 '16 at 02:38

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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 $f(x)<f(y)$.

Thomas Andrews
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  • This came to me because it answered a similar question I had a while back - what connected space $X$ has a group of self-automorphisms that is $1$-transitive on $X$, but not $2$-transitive on $X$. Brian Scott came up with the long line example here. http://math.stackexchange.com/questions/173385/connected-nice-topological-spaces-that-are-transitive-but-not-2-transitive-under – Thomas Andrews Feb 06 '16 at 02:43
  • (One-transitive is homogeneity, 2-transitive is stronger than your condition.) – Thomas Andrews Feb 06 '16 at 02:46