Region Less One Point is a 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 $R - \left\{{\zeta}\right\}$ 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 $R - \left\{{\zeta}\right\}$ 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 $R - \left\{{\zeta}\right\}$, and we are done.
If $\zeta \in \Gamma$, then we consider the $\epsilon$-neighborhood $N_\epsilon \left({\zeta}\right)$ of $\zeta$ for some $\epsilon$ such that $N_\epsilon \left({\zeta}\right) \subseteq R$.
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.
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