Closed Real Interval is Neighborhood Except at Endpoints

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\R$ be the real number line considered as an Euclidean space.

Let $\closedint a b \subset \R$ be a closed interval of $\R$.


Then $\closedint a b$ is a neighborhood of all of its points except $a$ and $b$.


Proof

Let $a < c < b$.

Let $\epsilon < \map \min {b - c, c - a}$.

From the definition of positive it follows that $\epsilon \in \R_{>0}$.


Let $\map {B_\epsilon} c = \openint {c - \epsilon} {c + \epsilon}$ be the open $\epsilon$-ball of $c$.

We have that $c + \epsilon < b$ and $a < c - \epsilon$.

Thus:

$\map {B_\epsilon} c \subseteq \closedint a b$


This article, or a section of it, needs explaining.
In particular: There needs to be a result linked to that proves that $\map {B_\epsilon} c$ is in fact a subset of $\closedint a b$.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining 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 {{Explain}} from the code.


It follows that, by definition, $\closedint a b$ is a neighborhood of $c$.


Now consider $a$.

Let $\epsilon \in \R_{>0}$.

We have that $a - \epsilon < a$ and so $\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$ does not lie entirely in $\closedint a b$.


Similarly, consider $b$.

Let $\epsilon \in \R_{>0}$.

We have that $b < b + \epsilon$ and so $\map {B_\epsilon} b = \openint {b - \epsilon} {b + \epsilon}$ does not lie entirely in $\closedint a b$.


It follows that, by definition, $\closedint a b$ is a neighborhood of neither $a$ nor $b$.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Closed_Real_Interval_is_Neighborhood_Except_at_Endpoints&oldid=541719"