Region Less One Point is Region

Theorem
Let $M = \left({A, d}\right)$ be a metric space.

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

Let $\zeta \in R$.

Then $R \setminus \left\{{\zeta}\right\}$ 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 Metric Space is Not Open.

Therefore $R \setminus \left\{{\zeta}\right\}$ 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 $\zeta \notin \Gamma$, then $\Gamma$ is also a path in $R \setminus \left\{{\zeta}\right\}$, and we are done.

If $\zeta \in \Gamma$, then we consider the open $\epsilon$-ball $B_\epsilon \left({\zeta}\right)$ of $\zeta$ for some $\epsilon$ such that $B_\epsilon \left({\zeta}\right) \subseteq R$.