Let $S$ be a subset of $\mathbb{R}$ such that $2018$ is an interior point of $S$ . Which of the followinf is(are) true?

Question

$(a)$ $S$ contains an interval.

$(b)$ There is a sequence in $S$ which does not converge to $2018$.

$(c)$ There is an element $y\in S , y\ne2018$ such that $y$ is also an interior point of $S$.

$(d)$ There is point $z\in S$ , such that $|z-2018|=0.002018$ .

Try:

for $(a)$ & $(c)$ , by definition of interior points , if $2018$ is an interior point of $S$ , $\exists$ $\epsilon>0$ such that $(2018-\epsilon,2018+\epsilon)$ is subset of $S$.

$\implies$ $S$ contains an interval and there are $\infty$ly many interior points of $S$ other than $2018$.

for $(b)$ let $S=[2018,2019]\cup \{\frac{1}{n} :n\in \mathbb{N}\}$

Now, let $<a_n>=<1/n>\to 0\ne 2018$

Question

How to think for option $(d)$ ?

Answer 1

Your proofs for (a) and (c) look fine. For (b), a single example does not constitute a proof. But note that there is no requirement that the sequence in $S$ have distinct terms, so you can just choose some $s\in S$ with $s\neq 2018$ (which exists since $2018$ is an interior point) and set $x_n=s$ for all $n$.

For (d), consider for instance the set $S=(2018-10^{-6},2018+10^{-6})$.