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It is clear to me that the intersection of a finite collection of open sets is open, also that a countable intersection of a collection of open sets is not always open.

But What can be said of a countable intersection of semi-closed intervals, e.g.

$ \bigcap_{n=1}^{\infty}(-1/n, 1]$

Thanks in advance.

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3 Answers

up vote 3 down vote accepted

The intersection of a countably infinite collection of half-open intervals of the same type may be empty, a half-open interval of that type, or a closed interval, which may be degenerate (a singleton) or not. Examples:

$$\begin{align*} &\bigcap_{n\in\Bbb Z^+}\left(0,\frac1n\right]=\varnothing\\ &\bigcap_{n\in\Bbb Z^+}\left(-1,\frac1n\right]=(-1,0]\\ &\bigcap_{n\in\Bbb Z^+}\left(-\frac1n,\frac1n\right]=\{0\}\\ &\bigcap_{n\in\Bbb Z^+}\left(-\frac1n,1+\frac1n\right]=[0,1] \end{align*}$$

If the intervals are $(a_n,b_n]$ for $n\in\Bbb Z^+$, let $a=\sup_na_n$ and $b=\inf_nb_n$; then

$$\bigcap_{n\in\Bbb Z^+}(a_n,b_n]=\begin{cases} [a,b],&\text{if }a\notin\{a_n:n\in\Bbb Z^+\}\\\\ (a,b],&\text{if }a\in\{a_n:n\in\Bbb Z^+\}\;. \end{cases}$$

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Great answer, thanks for your help. –  Mario Oct 17 '12 at 20:15
    
@Mario: You’re welcome! –  Brian M. Scott Oct 17 '12 at 20:16
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It must be an interval, because any intersection of convex sets is convex. By taking finite intersections we can assume without loss of generality that the intersection is decreasing, that is, $I_{n+1} \subset I_n$ for each $n$. The intersection $I$ can be a closed interval as in your example, where $I = [0,1]$, or it can be a half-open interval $(a,b]$. It cannot have the form $[a,b)$ or $(a,b)$ because the limit of the right endpoints of the $I_n$'s will be the right endpoint of $I$ and will be in $I$ (unless $I$ is empty, which is also possible.)

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In a general topological space, the notion of a semi-closed set is normally not defined and if it were, it probably wouldn't be very useful. The reason we have this notion for $\mathbb{R}$ is just the coincidental fact that the boundary of a connected set of $\mathbb{R}$ consists of at most 2 points.

Regarding your example: $\cap^\infty_{n=1}(-1/n,1]=[0,1]$.

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