How do I show that this set is compact and not compact at the same time? Given:
1.) For each $n \in \mathbb{N}, L_n$ is a line segment from $(0,0)$ to $(1, \frac{1}{n}) $
2.)$ L_\infty$ is a line segment from $(0,0)$ to $(1,0) $
3.) Both $L_n$ and $L_\infty$ are equipped with the subspace topology induced on $\mathbb{R}^2$
4.) $X= \bigcup_{n=1}^\infty L_n $, where $X$ is a topology with open sets $Y$ such that $Y$ is open in $X$ IFF $Y \bigcap L_n$ is open for all $n \in \mathbb{N}$
How do I show that $X$ is not compact in the defined topology but is compact with respect to the subspace topology induced by the usual topology on $\mathbb{R}^2$?
With that being said, what possible open subset $Y$ in $X$ is open with respect to the defined topology on X but not with respect to the subspace topology on X induced by the usual topology on $\mathbb{R}^2$?

In the "defined topology", $\{(1,1/n):n\in\Bbb N\}$ is a closed
discrete subset. Compact spaces cannot have closed
infinite discrete subsets.
