Ytukyg

I need a very simple example of a set of real numbers (if there is any) that is neither closed nor open, along with an explanation of why it is so.

$[0,1)$

It is not open because there is no $epsilon > 0$ such that $(0-epsilon,0+epsilon) subseteq [0,1)$.

It is not closed because $1$ is a limit point of the set which is not contained in it.

For a slightly more exotic example, the rationals, $mathbb{Q}$.

They are not open because any interval about a rational point $r$, $(r-epsilon,r+epsilon)$, contains an irrational point.

They are not closed because every irrational point is the limit of a sequence of rational points. If $s$ is irrational, consider the sequence $left{ dfrac{lfloor10^n srfloor}{10^n} right}.$

Let $A = {frac{1}{n} : n in mathbb{N}}$.

$A$ is not closed since $0$ is a limit point of $A$, but $0 notin A$.

$A$ is not open since every ball around any point contains a point in $mathbb{R} - A$.

Take $mathbb{R}$ with the finite complement topology - that is, the open sets are exactly those with finite complement. Then $[0,1]$ is neither open nor closed. It is not open since $mathbb{R}setminus [0,1]=(-infty,0) cup (1,infty)$ is not finite, and it is not closed since its complement, $(-infty,0) cup (1,infty)$, is not open, as just demonstrated.

The interval $left ( 0,1 right )$ as a subset of $mathbb{R}^{2}$, that is $left { left ( x,0 right ) in mathbb{R}^{2}: x in left ( 0,1 right )right }$ is neither open nor closed because none of its points are interior points and $left ( 1,0 right )$ is a limit point not in the set.

The rational numbers $mathbb{Q}$ are neither open nor closed. Recall a subset S of a metric space X space is open if every point x in S has an $epsilon$-neighborhood $N_{epsilon}(x)$ that is a proper subset of S. And a set is closed if(and only if) its complement is open. So $forall (q in mathbb{Q},r in [mathbb{R} setminus mathbb{Q}], epsilon > 0, lambda > 0)$ , $N_{epsilon}(q) cap [mathbb{R} setminusmathbb{Q}]neq emptyset$ and $N_{lambda}(r) cap mathbb{Q} neq emptyset$. So $mathbb{Q}$ is not open since every $epsilon$-neighborhood or a rational number contains irrationals. But its complement $ [mathbb{R} setminusmathbb{Q}]$, the set of irrational numbers, is also not open since no $epsilon$-neighborhoods or irrationals contain exclusively irrationals. But the complement of the rationals is not open, so $mathbb{Q}$ cannot be closed either. Therefore, the set of rationals is neither open nor closed.

Might I add: most of the previous examples are of sets that are both open and closed(like the half-open intervals of the real line [0,1) for example).