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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.

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$[0,1)$

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.

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  • $\begingroup$ One last Question, does the set has be connected? for example:- [1,2]U[3,4] ... what can we say about this set? $\endgroup$ – Monkey D. Luffy Aug 24 '12 at 0:36
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    $\begingroup$ @MonkeyD.Luffy A finite union of closed sets is closed, an arbitrary intersection of closed sets is closed. DeMorgan's Law gives you the analogous statements for open sets. $\endgroup$ – Eugene Shvarts Aug 24 '12 at 0:37
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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 $\left\{ \dfrac{\lfloor10^n s\rfloor}{10^n} \right\}.$

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  • $\begingroup$ $\mathbb Q$ is exotic? $\endgroup$ – celtschk Sep 2 '18 at 10:28
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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 $\mathbb{R} - A$.

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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.

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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.

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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.

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).

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