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Which set is not the intersection of countably many open sets?

(a) $\mathbb{Z}$

(b) $\mathbb{Q}$

(c) $[0,\infty)$

(d) the complement of $\mathbb{Q}$

(e) the complement of $\mathbb{Z}$

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What have you tried? –  TMM Oct 22 '12 at 8:24
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Asking questions you don't understand is not a good way to start to learn a topic. –  Chris Eagle Oct 22 '12 at 14:24
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Are we now OK with answering verbatim GRE questions? –  user641 Oct 22 '12 at 18:18

3 Answers 3

up vote 1 down vote accepted

It is easy to see that whenever $A$ is a countable set of real numbers, we can write its complement as the following countable intersection: $$\mathbb R\setminus A=\bigcap_{a\in A}\mathbb R\setminus\{a\}$$ Since deleting just one point from the real numbers results in an open set, we have immediately that $\mathbb{R\setminus Z, R\setminus Q}$ are both countable intersections of open sets.

One slightly more difficult theorem is that every closed set is a countable intersection of open sets. In this case, though, we only have a closed interval to deal with (one could notice that $\mathbb Z$ is closed, and use the above fact, if it is known, otherwise -- read ahead). For closed intervals we have that $$[a,b]=\bigcap_{n\in\mathbb N}\left(a-\frac1n,b+\frac1n\right)$$

In our case $b=\infty$ so this amounts to the intersection of the intervals $(-\frac1n,\infty)$. And so we have eliminated the uncountable sets from the possible answers.

So which one is it? Is it $\mathbb Z$ or $\mathbb Q$ that cannot be written as a countable intersection of open sets? We can prove directly that $\mathbb Q$ cannot be written as such intersection, as a consequence of Baire category theorem. However we can also just write the open sets which intersect into $\mathbb Z$, which is possibly simpler.

We can write $Z_n=\bigcup_{k\in\mathbb Z}(k-\frac1n,k+\frac1n)$ as an open set (a union of open intervals). Now we can verify that $$\bigcap_{n\in\mathbb N}Z_n=\mathbb Z$$ Clearly every $k\in\mathbb Z$ is in the intersection, and since $\mathbb Z$ is discrete every point not in $\mathbb Z$ would be excluded at some point (you may wish to give a rigorous argument here, if you're not sure about it).

So we have concluded that it has to be $\mathbb Q$ which cannot be written as an intersection of open sets, but why doesn't the same trick as we did here with $\mathbb Z$ would work? Well, $\mathbb Q$ is dense, taking $Q_n$ defined similarly to $Z_n$ would have to cover the entire real line, for every $n$. So the intersection is not $\mathbb Q$ but rather $\mathbb R$.


Generally speaking now, a countable intersection of open sets is known as a $G_\delta$ set, or $\Pi^0_2$ sets. These are Borel sets, and for Borel sets we can prove the following theorem:

Theorem: If $X\subseteq\mathbb R$ is a Borel set, then $|X|$ is either finite, countably infinite, or $2^{\aleph_0}$.

I won't give the proof here, but see above that we have generated examples where the intersection is of size continuum, and an example which is countable. To generate a countable intersection of size $k$ for any finite number $k$ simply do the following:

Let $(a,b)$ be an interval such that $(a,b)\cap\mathbb Z$ has exactly $k$ elements. Now $(a,b)\cap\bigcap_{n\in\mathbb Z} Z_n$ is a countable intersection of open sets, which is equal to $(a,b)\cap\mathbb Z$ and therefore has exactly $k$ elements.

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I wrote that last section because this question is tagged as [cardinals]. –  Asaf Karagila Oct 22 '12 at 13:52

Hint: It is not all that easy to prove that $\mathbb{Q}$ is not such an intersection. For details, please see this question.

But it is straightforward to show that all the others are a countable intersection of open sets. For example, $[0,\infty)$ is the intersection of the open sets $(-\frac{1}{n},\infty)$, where $n$ ranges over the positive integers.

So since this is a multiple choice question $\dots$.

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@KevinCarlson: Thanks for the edit. –  André Nicolas Oct 22 '12 at 8:27

$$ \mathbb Z=\bigcap_{m\in\mathbb N}\left(\bigcup_{n\in\mathbb Z}(n-\frac1m,n+\frac1m)\right), $$

$$ [0,\infty)=\bigcap_{n\in\mathbb N}(-\frac1n,\infty), $$ $$ \mathbb R\setminus\mathbb Z=\bigcap_{n\in\mathbb Z}\mathbb R\setminus\{n\}. $$ $$ \mathbb R\setminus\mathbb Q=\bigcap_{q\in\mathbb Q}\mathbb R\setminus\{q\}. $$ So the answer is (b).

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You wrote $\mathbb Q$ wrong. Note that $\frac12$ is not in it because it is not in the intersection defining $\frac13$. Similarly $\mathbb Z$ is written wrong. You should have written it as $\bigcup\bigcap$, but that is no longer a countable intersection of open sets. –  Asaf Karagila Oct 22 '12 at 9:00
    
Yes, I had typos in those two. Thanks! –  Martin Argerami Oct 22 '12 at 9:32
    
For every $m$, the union of $(q-\frac1m,q+\frac1m)$ is $\mathbb R$. So the intersection of countably many copies of $\mathbb R$ is $\mathbb R$. The fact is $\mathbb Q$ cannot be written as a countable intersection of open sets. No matter how hard you try. –  Asaf Karagila Oct 22 '12 at 9:35
    
Furthermore, if $\mathbb{R\setminus Z}=\bigcap_{n\in\mathbb Z}\mathbb R\setminus\{n\}$ why is $\mathbb {R\setminus Q}$ not the intersection $\bigcap_{q\in\mathbb Q}\mathbb R\setminus\{q\}$? –  Asaf Karagila Oct 22 '12 at 9:43
    
Yes, you are right. I realized about that while I was hiking a couple hours ago. –  Martin Argerami Oct 22 '12 at 14:28

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