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 $\mathbb{R}^2$?
 A: In the subspace topology $X$ is obviously bounded so we need only show it is closed in order for it to be compact.
To that end, let $p_n=(x_n, y_n)$ be a sequence in $X$ that converges to some limit $p\in\mathbb R^2$.
We must show that $x\in X$.
We have that $(y_n)$ is a convergent sequence in $[0,1]$.
If $\lim_{n\to\infty} y_n = 0$, it suffices to note that $0\leqslant x_n \leqslant 1$ to conclude that  $\lim_{n\to\infty} x_n \in [0,1]$ and hence $\lim_{n\to\infty}p_n \in L_{\infty} \subset X$.
If $y = \lim_{n\to\infty} y_n > 0$, then there is some $m \in\mathbb N$ such that $y_n \geqslant y/2$ for all $n\geqslant m$.
Let $k = \max\{n\in\mathbb N\,|\,\frac1n \geqslant y/2\}$.
Then for all $n\geqslant m$ we have $p_n \in \cup_{i=1}^k\,L_i$, which is a finite union of closed subsets, and hence closed.
It follows that $p \in \cup_{i=1}^k\,L_i \subset X$.

In the defined topology, let $B = B_{1/2}(0)$ be the open ball of radius $1/2$ centered on the origin and for $n\in\mathbb N\cup\{\infty\}$ consider sets $U_n = L_n\cup(B\cap X)$.
For each fixed $n\in\mathbb N\cup\{\infty\}$ we have $U_n\cap L_n = L_n$ and for $m\neq n$ we have $U_n\cap L_m = B\cap L_m$, so the $U_n$ are all open.
Can you see that no finite subcover of $\cup_{n=1}^{\infty} U_n$ will cover $X$?
A: Every open set in $\mathbb{R}^2$ that includes the point $(1,0)$ must also include an open ball around it, and so would include $(1,\frac{1}{n})$ for infinitely many $n$. In the defined topology on $X$, $L_n \setminus \{(0,0)\}$ is open for every $n$ (including $\infty$, to abuse notation slightly). Can you see how to turn these into an open cover of $X$ with no finite subcover?
A: Here's one way to produce a bad cover of $X$.  Take a small $\varepsilon$-Ball $B$ centered at $(0,0)$.  For $n\in\mathbb N\cup\{\infty\}$, define an open set in $X$ by $U_n=(B\cap X)\cup L_n$.  Can you see why $\{U_n:n\in\mathbb N\cup\{\infty\}\}$ is an infinite cover with no subcover (finite or otherwise)?
To show that $X$ is compact with respect to the subspace topology from $\mathbb R^2$, just show that it is bounded (obvious) and closed.  To show it is closed, it's probably easiest to show that it's complement is open.
For your last question, the set $U_\infty$ given above is open in the defined topology, but not in the standard topology.
A: In the "defined topology", $\{(1,1/n):n\in\Bbb N\}$ is a closed
discrete subset. Compact spaces cannot have closed
infinite discrete subsets.
