# Removal of an arbitrary point of the boundary of a closed and connected $A\subseteq\Bbb R^2$ so the new set remains connected

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?

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.

• 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 ! Jan 7, 2022 at 12:30
• 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 $\color{blue}{(:}$ ) ? Jan 7, 2022 at 19:22
• 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. Jan 8, 2022 at 17:30
• Oops someone had asked for an example of compact connected set with non-connected boundary and then deleted its question !! Jan 9, 2022 at 5:46