Let X be connected, compact, and Hausdorff. Cut points and separations

Question

Let $ X$ be a connected, compact, Hausdorff space containing at least two elements. Let $a$ be a cut point of $X$ and let $C$ and $K$ form a separation of $X-${$a$}. Prove that $C$ has at least one non cut point.

So far my thought process is as such.

Assume that $C-${$a$}$=:U$ contains no non cut points. Then $U-${$s$} is disconnected for all $s \in U$

I have tried to do this directly so I feel the contradiction is the way to go. But, I am having trouble getting traction on this one. Any help or hints would be appreciated.

Answer 1

The question is incorrect as stated, and here is a counterexample. Let's work in the usual topology on $\mathbb R^2$ and let $C=[0,1]^2\setminus \{(0,0)\}$ and $K=[-1,0]^2\setminus \{(0,0)\}$. These form a separation for $X=[0,1]^2\cup [-1,0]^2$ after removing the origin. Then neither $C$ nor $K$ has a cutpoint. (Clearly $X$ satisfies your conditions; it is Hausdorff, as a subspace of a Hausdorff space; it is compact, as a finite union of compact spaces; it is connected, by the plank lemma.)

Answer 2

Even the corrected version of your problem is incorrect, and here is a counterexample: Let $X$ be the space $$\Huge \infty \normalsize,$$ the wedge sum of two circles. Then the point in the middle $a$ is a cut point. Each component of $X-a$, however, is homeomorphic to an open interval and thus has only cut points.