Consider the set $\mathbb{Z}$ of integers,with the topology $\tau$ in which every set is closed if and only if it is empty or $\mathbb{Z}$ or finite. Then which of the following statements are true?
$\tau$ is a subspace topology induced by the usual topology on $\mathbb{R}$.
$\mathbb{Z}$ is compact in the topology $\tau$.
$\mathbb{Z}$ is Hausdorff in the topology $\tau$
Every infinite subset of $\mathbb{Z}$ is dence in the topology $\tau$.
I don't have any idea.
For 1: It is obvious that the subspace topology of $\mathbb{Z}$ induced by the "usual" one of $\mathbb{R}$ is discret.
For 2: A set $A \subseteq \mathbb{Z}$ is open in $(\mathbb{Z}, \tau)$ if its compliment is finite. Thus if you 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 $C_i \in \mathcal{C}$ wich $x_i \in C_i$. (this is of course true for any topological space with cofinite topology, nothing special about $\mathbb{Z}$ here)
For 3: Let $x,y \in \mathbb{Z}$. Suppose $\tau$ is Hausdorff. Then you would have $U,V$ open sets seperating $x$ from $y$. In particular $U \cap V = \emptyset$. Hence, by de Morgan, $U^c \cup V^c = \mathbb{Z}$. But the complements $U^c$ and $V^c$ are finite, by definition. Thus a condtradiction. (we only used that $\mathbb{Z}$ is infinite, so this is true for any infinite set with this particular topology)
For 4: By definition closures are closed sets. Moreover $A \subseteq \bar{A}$, $\bar{A}$ is the closure of $A$. But in $\tau$ closed sets are either finite or the whole space. So if $A$ is infinite, only $\bar{A} = \mathbb{Z}$ is possible. (again nothing but infiniteness of $\mathbb{Z}$ has been used).