Prove the following statement or find a counterexample:

Let $A\subseteq\Bbb R^2$ be a closed and connected set. Then,$\exists c\in\partial A$ s. t. $A\setminus\{c\}$ is still connected.

I think I found some counterexamples:

$x$ or $y$ axis or any other line in $\Bbb R^2,$ as well as graphs of unbounded continuous functions defined on an open interval $I\subseteq\Bbb R$ or graphs of continuous functions defined on the whole $\Bbb R.$

Question :

Is the unboundedness necessary for the statement not to hold?


1 Answer 1


I found the following answer in Whyburn's book Analytic topology in Chapter 3, Theorem 6.1 :

Every continuum has at least two non-cut points.

Where, (Chapter 1, 10.)

A compact connected set will be called a continuum

and, (Chapter 3, 1.)

If $M$ is a connected set and $p$ is a point of $M$ such that the set $M \setminus p$ is not connected, then $p$ will be called a cut point of $M$.

Now, if we take $A \subseteq \mathbb{R}^2$ closed, connected and bounded, $A$ is compact by the Heine-Borel theorem. So it is a continuum and it has two non-cut points which means that there exists (at least two) points $c$ such that $A \setminus c$ is connected. This solves the case $\mathring{A} = \emptyset$ which intuitively corresponds to curves in $\mathbb{R}^2$.

For the general case, I think that the theorem still holds because $\mathring{A} \neq \emptyset$ looks like a strong condition to me but I haven't been through yet.

  • 1
    $\begingroup$ Take a closed « ring » in $\mathbb{R^2}$, ie just a closed ball amputed from an inside closed ball for instance $\{ x \in \mathbb{R}^2, \ 1 \leqslant ||x|| \leqslant 2 \}$. It is a closed blounded set thus compact, it is obviously path-connected so connected and $A \setminus \mathring{A} = \partial A = S(0,1) \cup S(0,2)$ is the union of the 2 circles of radius 1 and 2 which is not connected ! $\endgroup$
    – hugo_panch
    Jan 7, 2022 at 12:30
  • $\begingroup$ But this just shows $\operatorname{Int}(A)$ doesn't correspond to curves in $\Bbb R^2,$ right (I mean, your comment, not the accepted answer $\color{blue}{(:}$ ) ? $\endgroup$
    – PinkyWay
    Jan 7, 2022 at 19:22
  • 1
    $\begingroup$ A small remark: The Theorem 6.1 you cite also needs the assumption that the space contains more than one point to begin with. Of course a singleton is connected and compact but does not contain two non-cut points. But I also do not see how your comment helps with what you are trying to prove. $\endgroup$ Jan 8, 2022 at 17:30
  • $\begingroup$ Oops someone had asked for an example of compact connected set with non-connected boundary and then deleted its question !! $\endgroup$
    – hugo_panch
    Jan 9, 2022 at 5:46

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