Connected topological space such that the removal of any of its points disconnects it into exactly $3$ connected components?

$\mathbb R$ has the property of being a connected space which is divided into $2$ connected components by the removal of any of its points.

I'm trying to generalize this property by constructing a connected (otherwise a $4$ elements set with the discrete topology will work) topological space such that removing any of its points leaves you with exactly $3$ connected components or by showing that such a space cannot exist.

Are there spaces with this property?
If so how well behaved can they be in terms of first and second countability and separation axioms?

• Further question: Can it be made $T_1$ and second-countable? Dec 21, 2016 at 23:42

Edit (7/2/2020): PatrickR in the comments below presents the following argument that this construction doesn't actually work.

Essentially, open sets in $$X$$ arise as saturated open sets with respect to the quotient map $$G\to X.$$ However, any saturated open set containing $$v$$ must contain the entire space $$G.$$ This implies that $$v$$ belongs to the closure of any subset $$Y\subseteq X.$$ In particular, since connected components are closed, any connected component of $$X\setminus\{p\}$$ (where $$p\neq v$$) must contain $$v.$$ Thus, it is not true that $$X\setminus\{p\}$$ will have 3 connected components unless $$p = v.$$

At the moment, I'm not sure how to remedy this issue. If there's a way to salvage this construction, I'd love to see it!

Another idea that I believe would work with some checking of details: start with the following self-similar graph $$G$$.

Notice that removing any half-open edge (including one endpoint and excluding the other) will leave you with three connected components. Choose some vertex $$v$$. Then if we want to remove a (half-open) edge, take the closed end of the edge to be the end which is "farther" from $$v$$. In this way, we have a consistent and well-defined way (at least, it could be made so with some cleaning up) to remove an edge which will disconnect the space into 3 connected components. Now, form the quotient space $$X = G/\sim$$, where $$x\sim y$$ if $$x$$ and $$y$$ are in the same half-open edge, and the point $$v$$ chosen above satisfies $$v\sim p$$ if and only if $$v = p$$. The removal of a point from this quotient is essentially equivalent to removing either the chosen vertex $$v$$ or a half open edge from the graph $$G$$, and so $$X\setminus\{x\}$$ should have three connected components for any $$x\in X$$.

This construction should generalize to give connected spaces which have $$n$$ connected components for any $$n\geq 2$$ upon removal of any point (take a graph like $$G$$ with $$n$$ edges coming from each vertex and perform an analogous quotient operation).

• That's one hell of a beautiful graphic and construction. +1. Dec 21, 2016 at 23:01
• This is, unfortunately, horribly not even T1.
– user98602
Dec 21, 2016 at 23:18
• @Stahl Does this example really work? Open sets in $X$ come from saturated open sets in $G$. A saturated open set that contains a particular half-open edge also contains all the half-open edges attached to it directly or indirectly away from $v$ (towards the leaves). In particular any saturated open set in $G$ containing $v$ must be the whole space. So the only open set containing $v$ is the whole of $X$. In particular, $v$ belongs to the closure of any subset $A\subseteq X$. Now this shows that removing a half-open edge different from $v$ does not split the space ... (continued) Jun 28, 2020 at 6:43
• into three connected components. Since any connected component of a space is always closed, it must contain $v$, there is only one connected component in the space with that one edge removed. The only case that gives three connected components is when removing $v$ itself. Jun 28, 2020 at 6:45
• @Stahl In general, if $A$ is a subspace of a topological space $X$, the closure of a a subset $B\subseteq A$ in $A$ is the intersection of $A$ with the closure of $B$ in $X$: $\operatorname{cl}_A(B)=\operatorname{cl}_X(B)\cap A$. So for a point of $A$, it is the same to say it is in the closure of $B$ wrt the topology of $X$ or wrt the topology of the subspace $A$. Jul 3, 2020 at 1:58

Let $$X$$ be the set of all finite sequences $$\langle x_0,\ldots,x_n\rangle$$ such that $$n\in\Bbb N$$, and $$x_k\in\Bbb R^+$$ for $$k=0,\ldots,n$$. For $$r>0$$ and $$x=\langle x_0,\ldots,x_n\rangle\in X$$ let $$B(x,r)$$ be the set of points $$y=\langle y_0,\ldots,y_m\rangle\in X$$ such that

• $$m\ge n$$;
• $$y_k=x_k$$ for $$k;
• $$|y_n-x_n|; and
• $$|y_k| for $$n.

Then $$\{B(x,r):x\in X\text{ and }0 is a base for a connected topology $$\tau$$ on $$X$$. In fact, for each $$\langle x_0,\ldots,x_n\rangle\in X$$ the subspace

$$\{\langle x_0,\ldots,x_{n-1},x\rangle\in X:x\in\Bbb R^+\}$$

is homeomorphic to $$\Bbb R^+$$ with the usual topology, so $$X$$ is even path connected.

Now let $$x=\langle x_0,\ldots,x_n\rangle\in X$$, and let $$Y=X\setminus\{x\}$$. Then $$Y$$ has the following three components:

• $$C_0(x)=\{\langle y_0,\ldots,y_m\rangle\in X:m\ge n\text{ and }y_n>x_n\text{ and }y_k=x_k\text{ for }k=0,\ldots,n-1\}$$

• $$C_1(x)=\{\langle y_0,\ldots,y_m\rangle\in X:m>n\text{ and }y_k=x_k\text{ for }k=0,\ldots,n\}$$

• $$C_2(x)=Y\setminus\big(C_0(x)\cup C_1(x)\big)$$

The intuitive idea is straightforward and is essentially the same as that of Kaj Hansen’s answer. We start with $$\Bbb R^+$$; points on it correspond to the sequences in $$X$$ of length $$1$$. To each point $$x\in\Bbb R^+$$ we attach another copy of $$[0,\to)=\{0\}\cup\Bbb R^+$$ by identifying the $$0$$ of $$[0,\to)$$ with $$x$$; the sequence $$\langle x,y\rangle\in X$$ then corresponds to the point $$y$$ on the copy of $$[0,\to)$$ attached to $$x$$. The topology is that induced by the so-called jungle river metric on $$\Bbb R^+\times\Bbb R^+$$.

Then we do it again: to each point $$\langle x,y\rangle$$ we attach a copy of $$[0,\to)$$ by identifying $$\langle x,y\rangle$$ with the $$0$$ of that copy of $$[0,\to)$$, and we extend the topology in an analogous fashion. We keep going to get sequences of arbitrary finite length. The space $$X$$ is the direct limit of the finite stages of this construction.

• So this is first- but not second-countable? Dec 21, 2016 at 23:50
• @Akiva: That’s correct. For example, $\{\langle x,1\rangle:x\in\Bbb R^+\}$ is a closed, discrete set, so second countability is out of the question. Dec 21, 2016 at 23:56

To each point $$(n, 0,0) \in [0, 2\pi] \times (0, 0) \subset \mathbb{R}^3$$, connect a segment $$L_n$$ that is orthogonal to the $$x$$-axis, has an endpoint at $$(n, 0,0)$$, and is at an angle of $$n$$ radians relative to the $$z$$-axis. Notice that removing any point along $$[0, 2\pi] \times (0, 0)$$ results in a space with $$3$$ connected components.

The problem is that the same cannot be said if we remove points along any of the segments $$L_n$$. To remedy this, we repeat the above process on each of the $$L_n$$, choosing the length of each new segment small enough to be disjoint with everything constructed thus far. Looking at a bijection between each $$L_n$$ and $$[0, 2\pi]$$ will tell us how to "angle" these segments as we attach them.

Removing any point on $$[0, 2\pi] \times (0,0)$$ or any point on any $$L_n$$ yields $$3$$ connected components. However, this is not true of any of the newly added segments, so we must now iterate the process yet again on each of them. Repeat ad infinitum, and let $$X$$ be the union of $$[0, 2\pi] \times (0,0)$$ and all segments from each iteration of the process. $$X$$ will have the desired property.

Of course, this is space is a weird, fractally mess, but it should work.

As Stahl points out in the comments below, we can generalize this idea by working in $$\mathbb{R}^\mathbb{R}$$ instead so that the removal of any point in the space yields arbitrarily many connected components (even as many as $$\aleph_{0}$$ or $$\aleph_{1}$$). In particular, at each point along each segment, we can now attach arbitrarily many new segments (in mutually orthogonal directions) so that any point on any segment is "far" from any other attached segment, making further iteration possible. Thus, working in $$\mathbb{R}^\mathbb{R}$$ eliminates the worry of having to "angle" to-be-attached segments as above: there will always be a previously-unused direction available.

• I believe the answer I've given will generalize readily, but I think you might be able to remedy your situation by moving into $\Bbb R^{\Bbb R}$ and have each axis that you need to attach stick out orthogonally into an unused component of $\Bbb R^{\Bbb R}$. Then you can attach finitely many segments at each point along any segment and have all segments far enough away from each other to not cause problems (since they'll all be sticking out in orthogonal directions from each other). Of course, this has the downside of needing to work in $\Bbb R^{\Bbb R}$. Dec 21, 2016 at 23:04
• Excellent! Passing to $\mathbb{R}^\mathbb{R}$ does the trick. Thanks a bunch for your response + your actual post; I enjoyed thinking about this question. May 5, 2017 at 4:06