Open Real Interval is Open Set

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Theorem

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

Let $\openint a b \subset \R$ be an open interval of $\R$.


Then $\openint a b$ is an open set of $\R$.


Corollary

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

Let $A := \openint a \infty \subset \R$ be an open interval of $\R$.

Let $B := \openint {-\infty} b \subset \R$ be an open interval of $\R$.


Then both $A$ and $B$ are open sets of $\R$.


Proof

Let $c \in \R$ such that $a < c < b$.

Let $\epsilon = \min \set {b - c, c - a}$.

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 $c$.

Let $y \in \map {B_\epsilon} c$.

Then:

$\size {c - y} < \epsilon $

implies:

$-\epsilon < c - y < \epsilon$

We have that:

\(\ds a\) \(=\) \(\ds c - \paren {c - a}\)
\(\ds \) \(\le\) \(\ds c - \epsilon\)
\(\ds \) \(<\) \(\ds y\)
\(\ds \) \(<\) \(\ds c + \epsilon\)
\(\ds \) \(\le\) \(\ds c + \paren {b - c}\)
\(\ds \) \(=\) \(\ds b\)

In other words:

$y \in \map {B_\epsilon} c \implies y \in \openint a b$

Thus:

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

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

The result follows by definition of open set.

$\blacksquare$


Sources

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