Open Unit Interval on Rational Number Space is Bounded but not Compact

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Theorem

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.


Then:

$\openint 0 1 \cap \Q$ is totally bounded but not compact

where $\openint 0 1$ is the open unit interval.


Proof

We first show that $\openint 0 1 \cap \Q$ is totally bounded.

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

By the Archimedean Property of $\R$:

$\exists n \in \N: \dfrac 1 n < \epsilon$

We pick the numbers $\dfrac i n \in \openint 0 1 \cap \Q$, where $i \in \N$ and $0 < i < n$.

Then for all $x \in \openint 0 1 \cap \Q$ and $x \ge \dfrac 1 n$:

\(\ds \inf_{0 \mathop < i \mathop < n} \map d {x, \frac i n}\) \(=\) \(\ds \inf_{0 \mathop < i \mathop < n} \size {x - \frac i n}\) Definition of Euclidean Metric on Real Number Line
\(\ds \) \(\le\) \(\ds \size {\frac {n x} n - \frac {\floor {n x} } n }\) $\floor {n x}$ is an integer strictly between $0$ and $n$
\(\ds \) \(<\) \(\ds \frac 1 n\) Real Number minus Floor
\(\ds \) \(<\) \(\ds \epsilon\)

For $x \le \dfrac 1 n$:

\(\ds \inf_{0 \mathop < i \mathop < n} \map d {x, \frac i n}\) \(=\) \(\ds \inf_{0 \mathop < i \mathop < n} \size {x - \frac i n}\) Definition of Euclidean Metric on Real Number Line
\(\ds \) \(\le\) \(\ds \frac 1 n - x\)
\(\ds \) \(\le\) \(\ds \frac 1 n\)
\(\ds \) \(<\) \(\ds \epsilon\)

This shows that $\openint 0 1 \cap \Q$ totally bounded.


From the Heine-Borel Theorem on a metric space, $\openint 0 1 \cap \Q$ is compact if and only if it is both totally bounded and complete.

From Rational Number Space is not Complete Metric Space it follows that $\openint 0 1 \cap \Q$ is not compact.

$\blacksquare$


Sources

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