‘flimsy’ spaces: removing any nn points results in disconnectedness

Consider the following property:

R is a connected space, but R{p} is disconnected for every pR.

S1 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 S1{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 R is 1-flimsy and S1 is 2-flimsy (as S1{}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 .
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 vw or wv, 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 nN (disregarding the trivial examples above). The removal of one point results in a finite n1-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)?


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 xyX, then X{x,y} has exactly two connected components. For this, we consider 3 open sets U1,U2,U3 such that (U1U2U3){x,y}c=X{x,y}, U1U2{x,y}c=U1U3{x,y}c=U2U3{x,y}c=, and i{1,2,3}, Ui{x,y}c. If u1U1{x,y}c and u2U2{x,y}c, then we can show X{u1,u2} is connected.

The second big step is to consider x,t,sX, three distinct points of a 2-flimsy space. We denote C1(t),C2(t) the two connected components of X{x,t} and C1(s),C2(s) the two connected components of X{x,s}. We suppose sC1(t) and tC1(s). Then D=C1(t)C1(s) is one of the two connected components of X{t,s}. In fact, the finite number of connected components implies C2(t){x} is connected, so the same goes for (C2(t){x})(C2(s){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 UVD=, (UV)D=D, and UD and VD, and if uUD and vVD, then we show X{u} or X{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{y}, a 2-flimsy space) is open and closed in X{x,t,s} and in X{y,t,s}, so it is open and closed in X{t,s}, which is not connected. So X is not a 3-flimsy space after all.

Source : Link , Question Author : Babelfish , Answer Author : Babelfish

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