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Consider the open cover $\mathscr{F}=\begin{Bmatrix}\pmatrix{0,5-\frac{1}{n}}:n\in\mathbb{N}\end{Bmatrix}$ for the set $(0,5)$.
Question:
Is $(0,5)=\bigcup^\infty_{n=1}\pmatrix{0,5-\frac{1}{n}}$? Why does $\mathscr{F}$ have no finite subcover?
$\mathscr{F}$ is an open cover of $(0,5)$; it is not a cover of $\{(0,5)\}$.
Yes, $(0,5)=\bigcup_{n=1}^\infty\left(0,5-\frac1n\right)$. On the one hand it’s clear that $\left(0,5-\frac1n\right)\subseteq(0,5)$ for each $n\in\Bbb Z^+$, so $\bigcup_{n=1}^\infty\left(0,5-\frac1n\right)\subseteq(0,5)$. On the other hand, if $x\in(0,5)$, then $x<5$, and $5-x>0$. Thus, there is a positive integer $k$ such that $\frac1k<5-x$ and hence $x<5-\frac1k$, so that $x\in\left(0,5-\frac1n\right)\subseteq\bigcup_{n=1}^\infty\left(0,5-\frac1n\right)$. Since $x$ was an arbitrary element of $(0,5)$, it follows that $(0,5)\subseteq\bigcup_{n=1}^\infty\left(0,5-\frac1n\right)$ and hence that $(0,5)=\bigcup_{n=1}^\infty\left(0,5-\frac1n\right)$.
Let $\mathscr{F}_0$ be any finite subset of $\mathscr{F}$, say $\mathscr{F}_0=\left\{\left(0,5-\frac1{n_k}\right):k=1,\ldots,m\right\}$. Let $n=\max\{n_1,\ldots,n_m\}$; then $5-\frac1{n_k}\le\frac1n$ for $k=1,\ldots,m$, so $$\left(0,5-\frac1{n_k}\right)\subseteq\left(0,5-\frac1n\right)$$ for $k=1,\ldots,m$. This implies that $$\bigcup_{k=1}^m\left(0,5-\frac1{n_k}\right)=\left(0,5-\frac1n\right)\subsetneqq(0,5)$$ and hence that $\mathscr{F}_0$ does not cover $(0,5)$. That is, no finite subset of $\mathscr{F}$ covers $(0,5)$, i.e., $\mathscr{F}$ is an open cover of $(0,5)$ with no finite subcover.
