Exercise about separation axioms and compactness

Question

Let $\tau$ be a family of subsets of $\mathbb{R}$ composed of the empty set and of every subset that consists of $\mathbb{R^2}$ minus a finite number of points and lines. (For example: the set obtained removing from $\mathbb{R^2}$ the points of the y-axis and the point (5,4) belongs to $\tau$)

(i) Decide if $(X,\tau)$ is $T_0,T_1,T_2$.

(ii)Decide if $C=${$(x,y): x^2+y^2 <1 $} is compact in $(\mathbb{R^2},\tau)$

Proof.: (i) I found it's $T_1$ because the points are closed sets. In fact, given a point {$x$}: $C_X$({$x$})$=\mathbb{R^2}\setminus${$x$}, which is an open set in $(X,\tau)$. Since it's $T_1$, it's also $T_0$.

(ii) If I take open cover of $C$ by removing every "line" parallel to the y-axis ($x=x_i$) in [-1,1], then I can't take a finite sub cover because this wouldn't cover all $C$

Is it correct?

Answer 1

The set $S$ is compact. Consider an open cover $\mathcal U=\{U_i\}$ of $S$. Assume none of them are empty. Pick $U_1$. This misses finitely many points $p_1,\cdots p_n$ and finitely many lines $l_1,\cdots l_d$ which intersect our set $S$. Since $U$ is a cover, pick $n$ sets $V_i\in \mathcal U$ to cover the points. For each line $l_i$, either

1) all of the $U_j$ are obtained by throwing away all of $l_i$ in which case $\mathcal U$ is not a cover so this case never happens.

or

2) one of the $U_j$ contains $l_i$ with maybe finitely many points removed, say $n_i$ points. In this case, using that $\mathcal U$ is a cover, find $n_i$ open sets in $\mathcal U$ to cover these points.

We now have a recipe for a finite cover which has no more than $n+\sum_{i=1}^d n_i$ sets of $\mathcal U$.

Answer 2

(i) is correct, but should argue why it's not T2. This is easy since there is not two disjointed sets in $\tau$. Assume there are two $U,V\in \tau:U\cap V=\emptyset$. Then $C=\mathbb{R}^2\setminus V\supseteq U$ so $U$ is in a set of finite points a lines wich implies that $\mathbb{R}^2=U\cup \mathbb{R}^2\setminus U$ is a finite union of points and lines.

(ii) It's compact. Let $\lbrace U_i\rbrace_{i\in I}$ be a open cover of $C$. Take some $i_0\in I$. Then there are finite points and lines of $C$ no covered by $U_{i_0}$. Take one open set in cover by point no covered, which must exists since is a cover. The idea is do the same for lines; so lets prove that given a open set $V$ and a line $L$, they're disjointed or the line is contained in the open set, except by finite points. Suppose $L\cap U^c\neq\emptyset$. Since $U^c$ is finite lines and points $L\subseteq U^c$ or $L\cap U^c$ are finite points. So since given some line no covered by $U_{i_0}$ that intersects $C$ it can't be in the complement of all open sets in cover, there is some that cover it except by finite points, which can be covered by other finite open set of the cover.