Vtyjkyuk

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, $mathbbQ$.

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 $left dfraclfloor10^n srfloor10^n right.$

Let $A = frac1n : n in mathbbN$.

$A$ 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 $mathbbR - A$.

Take $mathbbR$ 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 $mathbbRsetminus [0,1]=(-infty,0) cup (1,infty)$ 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 $mathbbR^2$, that is $left left ( x,0 right ) in mathbbR^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.