Beautiful Question of Real analysis, from IIT jam 2018
Question
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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