Is $\{(1/n, 2)\}$ an open cover of $(0, 1)?$

Question

I want to prove $C = \{(1/n, 2)\}$ for $n \in N$ is an open cover of $S = (0, 1).$

For any $n \in N, 1/n \in S,$ there exists $r > 0$ such that $(1/n - r, 1/n + r) \subset S,$ since $S$ is open. Take $0 < k < r.$ Then $0 < 1/n - r + k < 1/n,$ which is not covered by any $n,$ therefore $C$ is not a valid cover of $S.$

However, for any $x \in S,$ there exists an $N = 1/x$ such that $1/n < x < 2$ for any $n > N.$ In this way $C$ is an open cover of $S.$

Why?

Answer 1

"which is not covered by any $n$"

No.

It's not covered by THAT $n$. It is most certainly covered by a different natural number $m > \frac 1{1/n - r + k}$.

For any $x \in (0,1)$ then $0 < x$ and $\frac 1x > 0$. So there exists some $n > \frac 1x$ so $0 < \frac 1n < x < 1$ so $x \in (\frac 1n, 2)$.

Thus $\{(\frac 1n, 2)|n\in \mathbb N\}$ is an open cover. Period.

The fact that there is an $y$ so that $0 < y < \frac 1n < x < 1$ and $y \not \in (\frac 1n, 2)$ doesn't matter because there is an $m > \frac 1y > n > \frac 1x$ so that $y\in (\frac 1m,2)$.

This is why we define an open cover as a collection of open sets and not as a single open set.