Beautiful Question of Real analysis, from IIT jam 2018

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Solution: we discuss options one by one.

Recall that a point $p\in\mathbb{R}$ is said to be an interior point of $S\subseteq\mathbb{R}$ if 
i) $p\in S$
ii) there exists a neighborhood of $p$ which is contained in $S$.

Now given that, $2018\in S$ is an interior point of $S$. Hence by definition there exists a neighborhood of $2018$ which is contained in $S$, so that $S$ contains an interval. Hence (A) is true.

Now as $S\subseteq\mathbb{R}$, and $2018\in S$ is interior point of $S$ hence, it is an limit point of $S$ and hence there is sequence of points in $S$ which do not converge to $2018$. Hence (B) is also true.

Clearly (C) is also true by (A) (since $S$ contains a neighborhood (interval) containing $2018$ and we know by definition every point inside that interval is also an interior point of $S$ )

Finally for (D): we know by definition of an interior point $S$ contains a neighborhood (which is an interval) of point $2018$. Now this neighborhood can be too small such that, any point whose distance from $2018$ is greater or equal to $0.002018$ is not in that neighborhood. For example, if we take $0.00001$ neighborhood of $2018$ that is an interval $(2018-0.00001, 2018+0.00001)$ then it does not contain a point whose distance from $2018$ is $0.002018$ and hence $S$ not always, contain such points. Hence (D) is false too. 

Hence answer is (A),(B), (C) only.  

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