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Prove or contradict:

There are infinite open sets $U_1,U_2,...\in \mathbb{R}$ such that : $\mathbb{Q}=\bigcap^\infty _{i=1} U_i $

So I saw the following answer:

No, becuase if it was true,$\mathbb{Q}=\bigcap^\infty _{i=1} U_i $ ,then:

$\phi = \bigcap_{q\in \mathbb{Q}}\mathbb{R} \setminus \{q\} \ \cap \bigcap^\infty_{i=1}U_i$

which contradicts $Baire$ theorem.

Why does it contradicts it?

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    Can you state Baire's theorem? – 2017-02-09
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    @ErickWong For each countable collection of open dense sets, their intersection is dense. – 2017-02-09

2 Answers 2

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Baire's theorem says that the intersection of countably many open sets with a certain property is never empty.

  • What is that property?

  • Do you see why the sets of the form $\mathbb{R}\setminus\{q\}$ and $U_i$ have this property? (HINT: the first is easy; for the second, what can you say about the sets $U_i$ and $\mathbb{Q}$?)

Side note: "$\mathbb{R}\setminus q$" is incorrect notation. "$A\setminus B$" only makes sense if $A, B$ are both sets.

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    The first one is density? – 2017-02-09
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    @TakaTiki Exactly! So do you see how to show that each $\mathbb{R}\setminus\{q\}$ is dense? And do you see how to show that each $U_i$ is dense? – 2017-02-09
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    I understand why each $mathbb{R} \setminus \{q\}$ is dense,but why each $U_i$ is dense too? – 2017-02-09
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    @TakaTiki HINT: If $\mathbb{Q}=\bigcap U_i$, what can you say about $U_i$ versus $\mathbb{Q}$? – 2017-02-09
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    I'm sorry,I just can't figure it – 2017-02-09
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    @TakaTiki Alright, take a simpler example. Suppose $A=B\cap C$. What can you say about $A$ and $B$, for instance? *HINT: draw a [Venn diagram](https://en.wikipedia.org/wiki/Venn_diagram) . . .* – 2017-02-09
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    What do you mean "what can you say about $A$ and $B$"?I know that they both have the same certain properties that $C$ has? – 2017-02-09
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    @TakaTiki These two sets are related to each other in a certain way (and so are $A$ and $C$). Have you tried drawing a Venn diagram? A hint to get started: if $A$ has exactly 3 elements, could $B$ have exactly 2 elements? *And I don't know what you mean by "I know that they both have the same certain properties that $C$ has".* – 2017-02-09
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    So $|B|\ge|A|$? – 2017-02-09
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    @TakaTiki Yes, but you can in fact say a bit more. *Why* is $B$ at least as large as $A$? What do you know about elements of $A$, and elements of $B$? – 2017-02-09
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    Every element in $A$ must be in $B$ too – 2017-02-09
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    @TakaTiki **Exactly**! Or, to put it more concisely, $A\subseteq B$. Alright, now go back to the case we care about: $\mathbb{Q}=\bigcap_{i\in\mathbb{N}} U_i$, so what can you say about $\mathbb{Q}$ versus $U_i$? – 2017-02-09
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    That if $\mathbb{Q}$ is dense,then every $U_i$ is dense too? – 2017-02-09
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    @TakaTiki Yes, do you see why? – 2017-02-09
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    Not really,I can see that $\mathbb{Q}$ is subset of the intersectopn(and otherwise) but I'm not sure how does it help – 2017-02-09
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    @TakaTiki OK. Let's recap. You know that if $A=B\cap C$ then $A\subseteq B$. So - going from a finite intersection to an infinite intersection - if $\mathbb{Q}=\bigcap U_i$, then $\mathbb{Q}$???$U_i$. What should go in place of "???"? That is, what's the relationship between $\mathbb{Q}$ and $U_i$? – 2017-02-09
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    Well just like I said that $\mathbb{Q}$ is subset of each $U_i$ – 2017-02-09
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    @TakaTiki Yes. Now, do you know how to show that if $A$ is dense and $A\subseteq B$, then $B$ is also dense? HINT: let $V$ be a nonempty open set. Since $A$ is dense, $a\in V$ for some $a\in A$. Now can you find a $b\in B$ with $b\in V$? – 2017-02-09
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    The same $b=a$ because $A$ is a subset of $B$? – 2017-02-09
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    @TakaTiki Exactly! So do you see why this means that $B$ is also dense? – 2017-02-09
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    Indeed, I cant see it now.Thank you very much. – 2017-02-09
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$\mathbb{R}\setminus\{q\}$ is open as $\{q\}$ is closed, and dense as $\{q\}$ is not isolated.

If $\mathbb{Q} =\cap_n U_n$, where all $U_n$ are open, $\mathbb{Q} \subset U_n$ for all $n$, so every $U_n$ is dense as it contains a dense set: $\mathbb{R} =\overline{\mathbb{Q}} \subset \overline{U_n} \subset \mathbb{R}$.

The intersection of all these (countably many, as $\mathbb{Q}$ is countable) dense open sets is empty, so far from dense. This contradicts Baireness of the reals, which follows either from its local compactness or it's metric completeness, depending on your taste or background.