Suppose that A and B are two non-empty subsets of  $$\R$$ and, $$C=A \cap B$$ .

Which of the following conditions imply that C is empty . 

Correct option is A

Results :  

(i)  $$\textbf{Heine–Borel Theorem.:}$$  A non empty subset of R is compact if and only if it is both closed and bounded . 

(ii) Intersection of two non-empty open sets is always open .

Solution:

We know that union of two non-empty set is always open, so if A and B are two non-empty subsets of R 

such that. $$A \cap B=C$$​ then, C will be open (not closed) . and $$\textbf{cannot be compact.}$$ , So there does not exist such A and B.

Hence,

 $$\textbf{Option A is correct.}$$   

counter examples:

1.) For option B:

let A=B=R $$\implies A\cap B=\R\ (\text{Clopen, so can be considered as closed.})$$ 

2.) For option C :

 $$Let,\ A=\R \ and \ B=Q(\text{set of rationals)}\\Then,\ A\cap B= Q(non-empty)$$​​

3.) For option D:

Let A and B be intervals such that , A = (2,3) and B=[0,5] .Here A is open and B is compact But,

​$$A\cap B=(2,3) \text{ is non-empty.}$$​​