An example of neither open nor closed set 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.
 A: $[0,1)$
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.
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.
A: 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.   
A: 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 $\left\{ \dfrac{\lfloor10^n s\rfloor}{10^n} \right\}.$
A: 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 $\mathbb{R} - A$.
A: 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 $\forall (q \in \mathbb{Q},r \in [\mathbb{R} \setminus \mathbb{Q}], \epsilon > 0, \lambda > 0)$ , $N_{\epsilon}(q) \cap [\mathbb{R}\ \setminus\mathbb{Q}]\neq \emptyset$ 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}\ \setminus\mathbb{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.
