# Connectedness of the complements of the connected subsets

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)

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.
(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.
• 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. – Dylan Thurston May 9 '15 at 17:40