The set of rational numbers in the interval $(0,1)$ cannot be expressed as the intersection of a countable collection of open sets

I found this proof on a certain web page

A direct proof would be to show that if $A_k$ is a countable family of open sets whose intersection contains only the rationals then the intersection of the $A_k$ must contain an uncountable number of points. That's a contradiction.

Show you can assume the family is nested. Then the general idea is to show that for each of the open intervals in one of the sets $A_i$, there is an $A_n ~;n>i$ that contains at least two open intervals contained the original interval whose size is less than half the original interval.

So the intervals split endlessly. There are an uncountable numbers of ways of following this splitting down to a limit point.

Query: I understood the strategy and most of the part of this answer except the one in blue box. It says that for any open set $A_i$, we can find an $A_n$ such that $A_n$ contains at least two open intervals contained in $A_i$?

Attempt: I am trying to reason the above as follows :

None of the open sets in $\cap A_k$ can be disjoint from each other, else, the intersection will be the empty set contrary to our assumption.

This means, for any two open sets $A_i,A_j$ in the collection $\cap A_k : A_i \cap A_j \ne \{\phi \}$ .

$ A_i \cap A_j$ is an open set, hence, it contains an open interval $I$. This interval $I$ can be split endlessly and in uncountable ways.

But, whichever, split interval we get, it will always have a lot more numbers than the rationals.

Hence, any such interval must be uncountable contrary to our assumption.

Did I infer it correctly?

Please note that I am taking an introductory class in Real Analysis ( Point Set Topology) and haven't yet come across Baire's Category Theorem.

Thank you for your help..


This is a standard application of Baire's Theorem (BCT3):

Assume that indeed $$ (0,1)\cap\mathbb Q=\bigcap_{n\in\mathbb N}U_n, $$ where $U_n$ open dense in $[0,1]$, and $(0,1)\cap\mathbb Q=\{q_n\}_{n\in\mathbb N}$.

Now define $$ V_n=U_n\smallsetminus\{q_n\}. $$ Then clearly, $V_n$ open and dense in $[0,1]$, and $$ \bigcap_{n\in\mathbb N}V_n=\varnothing, $$ which contradicts Baire's Theorem, as $[0,1]$ is a complete metric spaces, and an intersection of countably many open and dense subsets of it is empty.

EDIT. If you do not want to use Baire's Theorem, let $U_n$'s and $V_n$'s defined as above. As $V_1$ is open, then there is a non-empty interval $(a_1,b_1)$, such that $$ (a_1,b_1)\subset V_1\subset (0,1). $$ Choose now a closed interval $[c_1,d_1]\subset (a_1,b_1)$, with $d_1>c_1$.

Next, as $V_2$ is open and dense, then there is a non-empty interval $(a_2,b_2)$, such that $$ (a_2,b_2)\subset V_2\cap (c_1,d_1), $$ and choose a closed interval $[c_2,d_2]\subset (a_2,b_2)$, with $d_2>c_2$.

Recursively we can thus obtain two sequences of intervals $(a_n,b_n)$, $[c_n,d_n]$, $n\in\mathbb N$, such that $$ [c_{n+1},b_{n+1}]\subset (a_{n+1},b_{n+1})\subset V_{n+1}\cap (c_n,d_n). $$ But $\bigcap_{n\in\mathbb N} V_n=\varnothing$ implies that $\bigcap_{n\in\mathbb N} [c_n,d_n]=\varnothing$, which is a contradiction.

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    $\begingroup$ Thank you for the answer. I am afraid though I haven't come across the Baire's Theorem. I am taking an introductory class in Point Set Topology . Could you please explain using the explanation given in the question? $\endgroup$ – MathMan Oct 1 '14 at 7:35
  • $\begingroup$ See the addition in my answer - it is basically the proof of Baire's Theorem. $\endgroup$ – Yiorgos S. Smyrlis Oct 1 '14 at 7:57
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    $\begingroup$ @YiorgosS.Smyrlis: Why do $U_{n}$ have to be dense? The statement only says that rationals in $(0,1)$ can not be expressed as a countable intersection of open sets, not necessarily dense ones. $\endgroup$ – T. Eskin Oct 2 '14 at 22:21
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    $\begingroup$ The rational can not be expressed as an intersection of countably many open set in $\mathbb R$. Hence, the rationals are a subset of each of these open sets, and thus they are all dense in $\mathbb R$. $\endgroup$ – Yiorgos S. Smyrlis Oct 2 '14 at 22:24
  • $\begingroup$ @YiorgosS.Smyrlis. Got it. Thanks. $\endgroup$ – T. Eskin Oct 2 '14 at 22:32

The rationals of (0,1) are meager. If they were $G_\delta$,
the irrationals of (0,1) would be $F_\sigma$. Each closed set
of the closed sets whose whose union is the irrationls of (0,1),
has empty interior. Thus they would be meager. Consequently
(0,1) would be meager, which it isn't.


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