16-03-2023
Problem: Show that each of the following families is an open cover of $(0,1)$.
- $\mathcal{C}_1 = \left\{ \left( 0,\frac{n-1}{n} \right) : n\in \mathbb{N} \right\} $;
- $\mathcal{C} _2 = \left\{ \left( \frac{1}{2}-\delta_n, \frac{1}{2} + \delta _n \right): \delta _n = \frac{n-2}{2n},~n\geq 3,~n\in \mathbb{N} \right\} $.
Solution: Note that all the elements of $\mathcal{C} _1$ and $\mathcal{C} _2$ are open. So we just need to prove that they cover the set $(0,1)$.
- We need to show that $(0,1)\subseteq \mathcal{C} _1$. Let $x \in (0,1)$ We need to show that $x\in \mathcal{C} _1$, that is, there exists $n_0\in \mathbb{N} $ such that $x\in \left( 0,\frac{n_0-1}{n_0} \right) $. Observe that \begin{align*} x\in \left( 0,\frac{n-1}{n} \right) & \implies 0 \lt x \lt \frac{n-1}{n} \\ & \implies 0 \lt x \lt 1 -\frac{1}{n} \\ & \implies 1-x > \frac{1}{n}. \end{align*} Since $x<1$ which implies $1-x > 0$ and hence by the Archimedean property we can find $n_0\in \mathbb{N} $ such that \[ 1-x > \frac{1}{n_0}. \] Therefore, $x\in \left( 0,\frac{n_0-1}{n_0} \right) $.
-
We need to show that $(0,1)\subseteq \mathcal{C} _2$. Let $x \in (0,1)$ We need to show that $x\in \mathcal{C} _2$, that is, there exists $n_0\in \mathbb{N} $ such that $x\in \left( \frac{1}{2}- \delta_{n_0},\frac{1}{2} + \delta _{n_0} \right) $. Observe that
\begin{align*}
\frac{1}{2} - \delta _n \lt x \lt \frac{1}{2} + \delta _n & \implies \frac{1}{2} - \frac{n-2}{2n} \lt x \lt \frac{1}{2} + \frac{n-2}{2n}
\\
& \implies \frac{1}{2} - \frac{n}{2n} + \frac{2}{2n} \lt x \lt \frac{1}{2} + \frac{n}{2n} - \frac{2}{2n}
\\
& \implies \frac{1}{2} - \frac{1}{2} + \frac{1}{n} \lt x \lt \frac{1}{2} + \frac{1}{2} - \frac{1}{n}
\\
& \implies \frac{1}{n} \lt x \lt 1-\frac{1}{n}.
\end{align*}
Therefore, we need to find $n_0\in \mathbb{N} $ such that
\[
\frac{1}{n_0} \lt x \lt 1- \frac{1}{n_0}.
\]
Note that
\begin{align*}
x \lt 1- \frac{1}{n_0} \implies 1-x> \frac{1}{n_0}.
\end{align*}
Thus, we want $n_0$ such that
\[
x > \frac{1}{n_0} \text{ and } 1-x > \frac{1}{n_0}.
\]
If $x \lt \frac{1}{2}$, then we have $0 \lt x \lt 1-x$. By the Archimedean property, we choose $n_0$ such that \[ \frac{1}{n_0} \lt x \lt 1-x. \] If $x > \frac{1}{2}$, then we have $0 \lt 1-x \lt x$. By the Archimedean property, we choose $n_0$ such that \[ \frac{1}{n_0} \lt 1-x \lt x. \] Thus, we found $n_0\in \mathbb{N}$ such that $\frac{1}{n_0} \lt x \lt 1-\frac{1}{n_0}$ and hence $x\in \mathcal{C} _2$. Hence, $\mathcal{C} _2$ is an open cover of $(0,1)$.