Region Less One Point is Region

Theorem

Let $M = \struct {A, d}$ be a dense-in-itself metric space.

Let $R \subseteq M$ be a region of $M$.

Let $x \in R$.

Then $R \setminus \set x$ is also a region of $M$.

Proof

From the definition, a region is a non-empty, open, path-connected subset of $M$.

First, note that as $R$ is open it can not be a singleton from Finite Subspace of Dense-in-itself Metric Space is not Open.

Therefore $R \setminus \set x$ is not empty.

Next, we see that from Open Set Less One Point is Open that $R$ is open.

Now, let $\alpha, \beta \in R$.

As $R$ is path-connected, we can join $\alpha$ and $\beta$ with a path $\Gamma$.

If $x \notin \Gamma$, then $\Gamma$ is also a path in $R \setminus \set x$, and the proof is complete.

If $x \in \Gamma$, then we consider the open $\epsilon$-ball $\map {B_\epsilon} x$ of $x$ for some $\epsilon$ such that $\map {B_\epsilon} x \subseteq R$.

Work In Progress In particular: Needs more work done to prove that a deleted neighborhood in a connected open set is path-connected. Obvious intuitively, but there's some groundwork to be done. Also see Euclidean Space without Origin is Path-Connected which is effectively the same result. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by completing it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{WIP}} from the code.