47
$\begingroup$

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)?

$\endgroup$
24
  • $\begingroup$ This is very similar to the concept of $n$ vertex connected in graph theory: en.wikipedia.org/wiki/K-vertex-connected_graph $\endgroup$ Oct 2 '18 at 14:36
  • $\begingroup$ @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. $\endgroup$
    – Babelfish
    Oct 2 '18 at 14:38
  • 1
    $\begingroup$ @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. $\endgroup$
    – MPW
    Oct 2 '18 at 14:44
  • 1
    $\begingroup$ @MPW you are right. I changed my definition accordingly. $\endgroup$
    – Babelfish
    Oct 2 '18 at 14:48
  • 3
    $\begingroup$ 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? $\endgroup$
    – Paul Frost
    Oct 3 '18 at 14:53
6
+200
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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.\\\... $\endgroup$
    – Babelfish
    Nov 29 '18 at 15:48
  • $\begingroup$ 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! $\endgroup$
    – Babelfish
    Nov 29 '18 at 15:50
5
+100
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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! $\endgroup$
    – Babelfish
    Oct 17 '18 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.