Let $\mathbb{R}$ with cofinite topology then infinite subset of $\mathbb{R}$ is? Question: let $τ_1$ be the usual topology on $\mathbb{R}$ and $τ_2$ be cofinite topology on $\mathbb{R}$ then $\mathbb{Z}$ is
(a) closed in $(\mathbb{R},τ_1)$ but not in $(\mathbb{R},τ_2)$
(b)closed in $(\mathbb{R},τ_2)$ but not in $(\mathbb{R},τ_1)$
(C) closed in both $(\mathbb{R},τ_1)$ and $(\mathbb{R},τ_2)$
(d) closed neither in $(\mathbb{R},τ_1)$ and $(\mathbb{R},τ_2)$
My attempt: we know under usual topology, $\mathbb{Z}$ is closed subset of $\mathbb{R}$ and hence $\mathbb{Z}$ is closed in $(\mathbb{R},τ_1)$. But, I am not sure about $(\mathbb{R},τ_2)$ please help me..
 A: This is one of those questions that is much easier to answer than it looks. As the various comments above suggest, you just need to take a deep breath and think carefully (and slowly!) about the definitions involved and then answer these questions: 
1) As has been mentioned above, you need to ask which sets are open, and which sets are closed. The definition of the co-finite topology tells you which sets are open. The definition of the co-finite topology also tells you which sets are open. (Hint: What's the definition of a closed set in terms of an open set?)  
2) Once you have the closed sets, you can ask whether they can possibly include Z. The comments above already give you the information you need to the answer to this question.
3) From your comments, it appears you may be thinking a set must be either open or closed. Is that true? Can a set be neither open nor closed? (Hint: Adding 0 to the open interval (0,1) gives the interval [0,1). Is this set open? Closed? Something else?)
A: Neither $\Bbb Z$ nor its complement $\Bbb R\setminus \Bbb Z$ is co-finite in $\Bbb R.$ So in the co-finite topology, neither $\Bbb Z$ nor its complement is open, so $\Bbb Z$ is neither open nor closed.
A: It is closed in usual topology.
Since, in the usual topology the set is closed as the set of limit points is empty. And in the other case your open sets are complements of finite set therefore , the set of limit points turns out to be the set of real number (It is dense in the second topology).
OR
Look at the complement of $\mathbb{Z} $ it's clearly not open as the only open sets  are complements of some finite subset of $\mathbb{R}$.
