# 'flimsy' spaces: removing any $n$ points results in disconnectedness

Consider the following property:

$$\mathbb R$$ is a connected space, but $$\mathbb R\setminus \{p\}$$ is disconnected for every $$p\in \mathbb R$$.

$$S^1$$ is a connected space and if we remove any point, it is still connected. But if we remove two arbitrary points $$p$$ and $$q$$, the resulting $$S^1 \setminus \{p,q\}$$ is disconnected.

Let $$X$$ be a topological space. Let's call $$X$$ to be $$n$$-flimsy if removing fewer then $$n$$ arbitrary points leaves the space connected and removing any $$n$$ arbitrary (distinct) points disconnects the space.

We saw that $$\mathbb R$$ is $$1$$-flimsy and $$S^1$$ is $$2$$-flimsy (as $$S^1 \setminus \{*\} \cong \mathbb R$$).

Question: Is there a $$3$$-flimsy space?

So I'm searching for a space $$X$$ such that the removal of any $$3$$ points disconnects the space, but fewer don't.

I suspect that there is no such space. I thought I could show it by showing first, that $$1$$- or $$2$$-flimsy spaces are in some way unique, but I found many examples of $$1$$-flimsy spaces which are significantly different (the long line, a variant of the topological sinus, trees).

Alternatively: Is there a standard terminology for this property? (it definitely 'feels' like $$n$$-connectivity in graph theory)

Addendum 1: A space $$X=\{x,y\}$$ with two points is a trivial $$3$$-flimsy example, since we cannot remove three distinct points. Of course, I'm interested in real examples.

Addendum 2: Since Qiaochu Yuan and Paul Frost argued that CW-complexes won't work, here are some thoughts concerning the finite case:

Let $$(X,T)$$ be a topological space with finite $$X$$. Then $$T$$ is automatically an Alexandrov topology and therefore has the Specialization preorder $$\prec$$. If we have a connected component $$Z(x)$$ of a point $$x$$ in a finite space with Alexandrov topology, then $$Z(x)$$ and its complement are closed and open, so they are downwardly closed. If we visualize $$(X,T)$$ by the graph $$G$$ which has $$X$$ as vertices and two vertices $$v,w$$ are connected if $$v\prec w$$ or $$w \prec v$$, then connected components in $$T$$ refer to connected components of the graph. Deleting a point in $$X$$ corresponds to deleting the respective vertex.

Claim: There is no finite $$1$$-flimsy space (disregarding the trivial examples above). Otherwise we have a graph where the removal of any vertex results in a disconnected graph. This graph can't be finite.

Corollary: There are no finite $$n$$-flimy spaces for $$n\in \mathbb N$$ (disregarding the trivial examples above). The removal of one point results in a finite $$n-1$$-flimsy space, which can't exist (induction).

Still open: Are there nontrivial $$3$$-flimsy spaces? Those should be infinite and shouldn't be homeomorphic to CW-complexes.

Addendum 3: Funfact: Every topological space can be embedded into a $$1$$-flimsy space. Just add a real line to each point (as a one-point union). Alternatively, add $$1$$-spheres to every point. Then add $$1$$-spheres to each new point. Continue like this for eternity.

Addendum 4: In the setting of Whyburn's book Analytic topology it is shown, that a compact set cannot be $$1$$-flimsy (Chapter 3, Theorem 6.1). Since all my examples for $$1$$-flimsy spaces are non-compact: Is there an example of a compact $$1$$-flimsy space? Are all $$n$$-flimsy spaces non-compact (at least they are infinite)?

• This is very similar to the concept of $n$ vertex connected in graph theory: en.wikipedia.org/wiki/K-vertex-connected_graph Oct 2, 2018 at 14:36
• @Lorenzo Yes, it definitely feels like it (as I just edited) and the standard $n=1,2$ examples are graphs, but for $n=3$ it seems to be much harder. Oct 2, 2018 at 14:38
• @Lorenzo: "Not 2-flimsy" doesn't necessarily mean $X\setminus\{p,q\}$ is always connected, just that there exists a pair $(p,q)$ for which it is connected (i.e., not always disconnected). In other words, your condition is stronger, I think.
– MPW
Oct 2, 2018 at 14:44
• @MPW you are right. I changed my definition accordingly. Oct 2, 2018 at 14:48
• Topological manifolds (with or without boundary) are not $3$-flimsy. As Qiaochu Yuan remarked, CW-complexes are not $3$-flimsy: CW-complexes of dimension $> 1$ are not because you can remove $3$ points of any open cell of dimension $> 1$ without disconnecting the space, and $1$-dimensional CW-complexes are not because removing two points from an open $1$-cell disconnects the space. Have you tried finite spaces with non-discrete topology? Oct 3, 2018 at 14:53

If I did not make any mistake, 3-flimsy spaces does not exist. You can check this link for my proof and some other results about 2-flimsy spaces. Without giving all the details, here are the big steps of the proof:

First, we show that if $$X$$ is a 2-flimsy space and $$x\neq y\in X$$, then $$X\backslash\{x,y\}$$ has exactly two connected components. For this, we consider 3 open sets $$U_1,U_2,U_3$$ such that $$(U_1\cup U_2\cup U_3)\cap\{x,y\}^{c}=X\backslash\{x,y\}$$, $$U_1\cap U_2\cap\{x,y\}^{c}=U_1\cap U_3\cap\{x,y\}^{c}=U_2\cap U_3\cap\{x,y\}^{c}=\emptyset$$, and $$\forall i\in\{1,2,3\},\ U_i\cap\{x,y\}^{c}\neq\emptyset$$. If $$u_1\in U_1\cap\{x,y\}^{c}$$ and $$u_2\in U_2\cap\{x,y\}^{c}$$, then we can show $$X\backslash\{u_1,u_2\}$$ is connected.

The second big step is to consider $$x,t,s\in X$$, three distinct points of a $$2$$-flimsy space. We denote $$C_1(t),C_2(t)$$ the two connected components of $$X\backslash\{x,t\}$$ and $$C_1(s),C_2(s)$$ the two connected components of $$X\backslash\{x,s\}$$. We suppose $$s\in C_1(t)$$ and $$t\in C_1(s)$$. Then $$D=C_1(t)\cap C_1(s)$$ is one of the two connected components of $$X\backslash\{t,s\}$$. In fact, the finite number of connected components implies $$C_2(t)\cup\{x\}$$ is connected, so the same goes for $$(C_2(t)\cup\{x\})\cup(C_2(s)\cup\{x\})$$ : the only thing to verify is the connectedness of $$D$$. The proof looks like to the first step. If $$U,V$$ are two open sets of $$X$$ such that $$U\cap V\cap D=\emptyset$$, $$(U\cup V)\cap D=D$$, and $$U\cap D\neq\emptyset$$ and $$V\cap D\neq\emptyset$$, and if $$u\in U\cap D$$ and $$v\in V\cap D$$, then we show $$X\backslash\{u\}$$ or $$X\backslash\{v\}$$ is not connected.

Finally, if $$X$$ is a $$3$$-flimsy space and $$x,y,t,s$$ some distinct points of $$X$$, then $$D$$ (defined as previously in $$X\backslash\{y\}$$, a 2-flimsy space) is open and closed in $$X\backslash\{x,t,s\}$$ and in $$X\backslash\{y,t,s\}$$, so it is open and closed in $$X\backslash\{t,s\}$$, which is not connected. So $$X$$ is not a 3-flimsy space after all.

• Hello Robin Khanfir, I finally got the time to parse your proof. It is not easy to read, but I think the proof is sound. Thank you very much for your efforts! I'm sorry it took me so long to proof-read.\\\ two minor typos: In the proof of Lemma 1, it should be $\tilde{U} \cup \tilde{V} \subset X \setminus \{u,v\}$ instead of $\tilde{U} \cap \tilde{V} \subset X \setminus \{u,v\}$. In Theorem 1, you write $U_2 \cap U_2$, which should be $U_2 \cap U_3$ instead.\\\... Nov 29, 2018 at 15:48
• I didn't quite get you argumentation in Theorem 1, when you write "either open in $X\setminus\{y\}$ and $X\setminus \{z\}$ or closed in $X\setminus \{y\}$ and $X \setminus\{z\}$, because $X\setminus\{y\}$ and $X \setminus\{z\}$ are n-flimsy". Though I checked the correctness myself, so it should be fine anyway. Thanks again! Nov 29, 2018 at 15:50

Here is a proposition that I believe will help to at least figure out whether or not a $$3$$-path-flimsy space exists. An $$n$$-path-flimsy space would be a space such that removing fewer than $$n$$ points would keep the space path-connected, but removing any $$n$$ points would make the space not path-connected.

Proposition A: Let $$X$$ be a $$2$$-path-flimsy space and $$x\in X$$. Then for any path-connected open neighborhood $$N$$ of $$x$$, such that $$X\setminus N$$ is also path-connected, the space $$N\setminus\{x\}$$ has at most two path-connected components.

Proof of Proposition A: The proof is by contradiction. Assume for the contrary that there exists $$x\in X$$ with a path-connected open neighborhood $$N$$, such that $$X\setminus N$$ is also path-connected, and such that the space $$N\setminus\{x\}$$ has three distinct path-connected components $$C_1$$, $$C_2$$ and $$C_3$$. Let $$c_i\in C_i$$. Since $$X$$ is $$2$$-path-flimsy, the space $$X\setminus\{x\}$$ is path-connected, so $$N\neq X$$, so we can find $$p\in X\setminus N$$.

Fix some $$1\leq i\leq3$$. Since $$N$$ is path-connected, it follows that the set $$C_i\cup\{x\}$$ is path-connected. This is because there is a path from $$x$$ to $$c_i$$ in $$N$$, and we can deduce that the last moment the path was not in $$C_i$$, it must have been at $$x$$ by the definition of path-connected component. By similar reasoning, we find that $$C_i\cup(X\setminus N)$$ is path-connected.

Since $$C_i$$ is a path-connected component of $$N\setminus\{x\}$$, any path that leaves $$C_i$$ must pass through $$X\setminus(N\setminus\{x\})=\{x\}\cup(X\setminus N)$$ first. Since $$X$$ is $$2$$-path-flimsy, $$X\setminus\{c_i\}$$ is path-connected, so from any $$c\in C_i$$ there is a path that leaves $$C_i$$. We can conclude that there is either a path from $$c$$ to $$x$$ in $$C_i\cup\{x\}$$, or there is a path from $$c$$ to $$p$$ in $$C_i\cup(X\setminus N)$$.

We can now conclude that $$X\setminus\{c_1,c_2\}$$ is path-connected, which contradicts the fact that $$X$$ is $$2$$-path-flimsy and finishes the proof of Proposition A. This is because every point is either path-connected to $$x$$ or $$p$$ and $$c_3$$ is path-connected to both.

• Somehow $n$-path-flimsy seems to be a feature that is easier to grasp than $n$-flimsy. In particular, some examples for $1$- or $2$-flimsy spaces are not $1$- or $2$-path-flimsy, like the sinus of the topologist./// Your proof is nice, I think I got it. And I think I see how your proof might show that there are no hausdorff $3$-path-flimsy spaces. Nice! Oct 17, 2018 at 14:46