EDIT: My original foolish version was instantly destroyed by Dylan Thurston; it consisted of questions 1 & 2 below. Thus now only new question 0 remains to be answered.

Let $\ X:=M^n\ $ be a connected manifold (with or without boundary) of dimension $\ n\ge 1,\ $ or let $\ X:=S^1\ $ be a circle. Then

$(E)\qquad X\setminus\{p\}\ $ is connected for every $\ p\in X$

Only other hand, of the mentioned spaces only the circle has the stronger property

$(B)\qquad X\setminus Y\ $ is connected for every connected subspace $\ Y\subseteq X$.

Thus $\ (B)\Rightarrow (E).\ $ Now, given a naturally or classically defined class of topological spaces which already have property $(E)$ one would like to prove property $(B)$. In particular, let me propose the following two conjectures (and their unspoken about obvious variations; the first conjecture is most likely already know, in which case I would like to ask about references):

0: Let $\ X\ $ be an arbitrary Hausdorff compact space such that for every open $\ G\subseteq X\times X,\ $ for which $\ \forall_{x\in X}(x\ x)\in G,\ $ there exists a continuous surjection $\ f:X\rightarrow S^1\ $ such that $\ \forall_{s\in S^1} f^{-1}(s)\times f^{-1}(s)\subseteq G.\ $ Does $\ X\ $ has property $(B)\ $ ?   Are such spaces the only ones among Hausdorff connected spaces, which have at least 2 different points, which have property $(E)\ $ ?

The two questions below were answered (immediately) by Dylan Thurston:

1:   Let $\ X\ $ be an arbitrary connected topological graph (the body of any finite 1-dimensional simplicial complex) which has no end-points (i.e. $(E)$ holds). Does $\ X\ $ have property $(B)$? (answered by Dylan Thurston)

2:   Let $\ X\ $ be an arbitrary connected 1-dimensional topological space, which has property $(E)$. Does $\ X\ $ have property $(B)$? (answered by Dylan Thurston)

About the topological dimension:

Here one may consider one of the three classical topological dimensions: $\dim$, ind, or Ind. One may consider separable metric spaces (when the three dimensions are equivalent), or Hausdorf compact spaces for the covering dimension $\dim$. Other variations are possible, interesting and welcome.

REMARK 1   For every Hausdorff connected compact space $\ X,\ $ which has at least two different points, there are at least two different points $\ p\in X\ $ such that $\ X\setminus\{p\}\ $ is connected.

REMARK 2   Property $(B)$ discussed in this post is antipodal to the notion of the biconnected spaces.


1 Answer 1


(B) does not hold for topological graphs. Let $X$ be the 1-skeleton of a tetrahedron, and let $S$ be a cycle of 4 edges. Then $X \setminus S$ is not connected.

  • 1
    $\begingroup$ In fact, (E) also does not hold for topological graphs without endpoints. Consider the "dumbbell graph" $D$ formed by taking two vertices $x$ and $y$, joining them by an edge, and attaching loops at both $x$ and $y$. (This is a CW complex, and you can subdivide to make this a graph in some more strict sense.) Then $D \setminus \{x\}$ is not connected. $\endgroup$ May 9, 2015 at 17:40
  • $\begingroup$ Ouch! Dylan, thank you. Obviously, you've answered both question since the answer to the second one, under the circumstances, follows instantly from yours. My initial reflex was that spaces with property (B) should be something like circle-like. I'll keep the previous version for the record, and may add a better version. $\endgroup$ May 9, 2015 at 17:48

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