Does $\bigcup_{n=1}^\infty \left(-\infty, 1-\frac{1}{n}\right) = (-\infty, 1)$? I'm currently trying to think of an example of a proper subset of $\mathbb{R}$ that is not compact in the topological space $(\mathbb{R}, \tau)$, where $\tau = \{(-\infty, a): a\in \mathbb{R}\}\cup\{\emptyset, \mathbb{R}\}$.
I suspect that $(-\infty, 1)$ is not compact. In trying to show this, I'm hoping that an open cover can be given by $\bigcup_{n=1}^\infty \left(-\infty, 1-\frac{1}{n}\right)$, but this is only true if $(-\infty, 1) \subset \bigcup_{n=1}^\infty \left(-\infty, 1-\frac{1}{n}\right)$. If this is true, then certainly there is no finite subcover of $(-\infty, 1)$ coming from $\bigcup_{n=1}^\infty \left(-\infty, 1-\frac{1}{n}\right)$, and so $(-\infty, 1)$ is not compact.
So, does $\bigcup_{n=1}^\infty (-\infty, 1-\frac{1}{n}) = (-\infty, 1)$?

We wish to show that for any $x \in (-\infty, 1)$, there exists a positive integer $n$ such that $x < 1 - 1/n$, or equivalently, $$n > 1/(1-x).$$  Recall the Archimedean property of the reals, which states that for any reals $a, b > 0$, there exists a positive integer $n$ such that $na > b$.  So if we choose $$a = 1, \quad b = 1/(1-x),$$ (the latter choice being positive since $x < 1$ implies $1-x > 0$), we obtain the desired assertion, and the proof is complete.
