Open Real Interval is not Compact

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Theorem

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

Let $I = \openint a b$ be an open real interval.


Then $I$ is not compact.


Proof 1

From Open Real Interval is not Closed Set, $I$ is not a closed set of $\R$.

The result follows by definition of compact.

$\blacksquare$


Proof 2

It suffices to demonstrate this for a particular open interval: we use $\openint 0 1$.

Consider the set of open intervals $\openint {\dfrac 1 n} 1$ for all $n \in \Z_{>1}$.

Each of these is an open set in $\openint 0 1$.

Also:

$\openint 0 1 = \ds \bigcup_{n \mathop \ge 2} \openint {\dfrac 1 n} 1$

Thus $\ds \bigcup_{n \mathop \ge 2} \openint {\dfrac 1 n} 1$ is an open cover for $\openint 0 1$.

Note that:

$\forall x \in \openint 0 1: \exists n \in \Z: n > \dfrac 1 x$

and so:

$x \in \openint {\dfrac 1 n} 1$

Consider a finite subcover of $\ds \bigcup_{n \mathop \ge 2} \openint {\dfrac 1 n} 1$.

It will be in the form:

$\set {\openint {\dfrac 1 {n_1} } 1, \openint {\dfrac 1 {n_2} } 1, \ldots, \openint {\dfrac 1 {n_r} } 1}$

But this is just $\openint {\dfrac 1 N} 1$, where $N = \max \set {n_1, n_2, \ldots, n_r}$.

So no finite subcover of $\ds \bigcup_{n \mathop \ge 2} \openint {\dfrac 1 n} 1$ can be a cover for $\openint 0 1$.

$\blacksquare$


Proof 3

We have that $\struct {\R, \size {\, \cdot \,}}$ is a normed vector space.

Also, $I$ is bounded by $a$ and $b$.

Consider a sequence $\ds \sequence {x_n}_{n \mathop \in \N}$ with $\ds x_n = a + \frac {b - a} {2n}$.

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

Let $\ds N = \frac {b - a} {2 \epsilon}$.

Then for all $n \in \N$ such that $n > N$ we have that:

\(\ds \size {a + \frac {b - a} {2n} - a}\) \(=\) \(\ds \size {\frac {b - a} {2n} }\)
\(\ds \) \(<\) \(\ds \frac {b - a} {2N}\) $b > a$
\(\ds \) \(=\) \(\ds \frac {b - a} 2 \frac {2 \epsilon} {b - a}\)
\(\ds \) \(=\) \(\ds \epsilon\)

Hence, $\sequence {x_n}_{n \mathop \in \N}$ converges to $a$.

By Limit of Subsequence equals Limit of Real Sequence, every subsequence of $\sequence {x_n}_{n \mathop \in \N}$ converges to $a$ as well.

However, $a \notin I$.

By definition, $I$ is not compact.

$\blacksquare$


Also see


Sources

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