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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. And also, a very short and simple explanation of why it is neither closed nor open.
Thank you!!

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Why did you delete your previous question? – Andrés E. Caicedo Aug 24 '12 at 0:29
Because people keep downvoting without trying to understand what I'm trying to say. – Monkey D. Luffy Aug 24 '12 at 0:31
I was writing a comment with some advise, on the issues you seemed to be having with the notation, which I believe was the source of the difficulties. The question disappeared as I was posting it. Oh, well... – Andrés E. Caicedo Aug 24 '12 at 0:34
@AndresCaicedo I'm terribly sorry ... I just needed to check that those two sets are "neither open not closed sets" – Monkey D. Luffy Aug 24 '12 at 0:38
@AndresCaicedo I would be happy to receive your advise even now!! – Monkey D. Luffy Aug 24 '12 at 0:39
up vote 15 down vote accepted


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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One last Question, does the set has be connected? for example:- [1,2]U[3,4] ... what can we say about this set? – Monkey D. Luffy Aug 24 '12 at 0:36
@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. – Eugene Shvarts Aug 24 '12 at 0:37
all right ... thanks!! – Monkey D. Luffy Aug 24 '12 at 0:43

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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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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