Consider the topology $T = \{U \subseteq \mathbb{Z} : \mathbb{Z} \setminus U \ \text {is finite or} \ 0 \notin U\}$ on $\mathbb{Z}$ ,then the topological space $(\mathbb{Z},T)$ is
choose the correct option
a) compact but Not connected
b) connected but not compact
c)both compact and connected
d) neither compact nor connected
My attempt : my answer is option C
For Compactness : set $U \subseteq \mathbb{Z}$ is open in $(\mathbb{Z}, T)$ if its compliment is finite. Thus if we have an open cover $\mathcal{C}$ take any $C \in \mathcal{C}$. Since $C$ is open it covers all of $\mathbb{Z}$ except some finite number of points, say $x_1, \dots, x_n \in \mathbb{Z}$. Take $C_i \in \mathcal{C}$ which $x_i \in C_i$. so $(\mathbb{Z},T)$ is compact
For connnectedness : Let $x,y \in \mathbb{Z}$. Suppose $T $ is disconnected Then we would have $P,Q$ open sets separating $x$ from $y$. In particular $P \cap Q= \emptyset$. Hence, by de Morgan, $P^c \cup Q^c = \mathbb{Z}$. But the complements $P^c$ and $Q^c$ are finite, by definition. Thus a condtradiction. because we know that $\mathbb{Z}$ is infinite, So $(\mathbb{Z},T)$ must be connected
is it correct ??
Pliz verified if im wrong
any hints/solution will be appreciated
thanks in advance
Yes, option c) is the correct one.
When you proved that it is compact, you should have added that $\{C,C_1,C_2,\ldots,C_n\}$ is a finite subcover of $\mathcal C$.
And, concerning the proof of connectedness, you should have added that neither $P$ nor $Q$ is equal to $\emptyset$ or to $\mathbb Z$ (a fact that you used implicitly).