Open cover, finite subcover 
Find an open cover of $(-\infty,0]$ that does not have a finite subcover.

Can anyone help me with this? Please explain the intuition of finding it too.
Thank you so much!
 A: A very easy solution is just $$\bigcup_{n=1}^\infty(-n,0].$$ Clearly it can't be finite, and it's open.
If the space isn't to be taken as $(-\infty,0]$, then take the union to be $$\bigcup_{n=1}^\infty(-n,1).$$
A: For every rational number $r \in A = (-\infty, 0]$ take $\{B(1,r)\}_{r \in A}$. Given any $x \in A$, we know by density of rationals that there exists $r_x$ such that $|r_x - x|< \epsilon<1 \Rightarrow x \in B(1,r_x)$. However, if we suppose $\{B(1,r_i)\}_{i =1}^n$ is a finite sub cover. Then if we take $s < \min\{r_i\} - 2$, so $s \in A \setminus \bigcup_i B(1,r_i)$.
A: One way of going about this is to find an open cover where, if any sets are missing, it clearly dons't cover the set.
For example, consider $\{(k-3/4,k+3/4):k\in\Bbb{Z}\}$. In other words, we put a cover around each integer with length $1.5$ (to make sure they overlap). If any of these sets $(k-3/4,k+3/4)$ is missing, then $k$ is not covered. Thus we always will need each $k\in\Bbb{Z}$ where $k\le 0$.
A: There are basically two ways to do this.
(1) infinitely many intervals but a finite number of centers.
For example, all intervals with finite radius and center 0.
(2) infinitely many intervals but with bounded radii.
For example, all intervals with center in $(-\infty,0]$ and radius 1.
A: Maybe you want $C = \{(-n-2, -n+2), n \in \mathbb{N}\}$
Let $V_n = (-n-2, -n+2)$
Then $[0, \infty) \subseteq \bigcup\limits_{n = 1}^\infty V_n $
Try to find an finite subcover $U$ such that $[0, \infty)  \subseteq U$, you can't!
