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If $\{T_n\}$ is a sequence of sets that converges to the set of irrational numbers such that $T_1 \subseteq T_2 \subseteq T_3 \subseteq \ldots$. Must $\overline{T_n}$ contain an interval for some $n$? $\overline{T_n}$ is the closure of $T_n$. I am asking must there exist some $a, b \in \mathbb{R}$ such that $(a,b) \subset \overline{T_n}$ if $n$ is large enough.

I believe it to be true since I can't find a counterexample, but I can't seem to prove it . Does it have to do with the completeness of real numbers?

By convergence, I mean that for every irrational number $x \in ℝ$, there exists $N$, such that $x \in T_n$ for all $n \geq N$.

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  • $\begingroup$ In what sense do you mean "converge"? $\endgroup$ – Lucas Mann Sep 25 '14 at 15:32
  • $\begingroup$ It's not clear what do you mean by saying that a sequence of sets is convergent. $\endgroup$ – Crostul Sep 25 '14 at 15:33
  • $\begingroup$ For every irrational number $x \in \mathbb{R}$, there exists $N$, such that $x \in T_n$ for all $n \geq N$. $\endgroup$ – ppham27 Sep 25 '14 at 15:34
  • $\begingroup$ So, the sequence $\:\mathbb{R},\hspace{-0.03 in}\mathbb{R},\hspace{-0.03 in}\mathbb{R},...\:$ converges to the set of irrational numbers? $\;\;\;\;$ $\endgroup$ – user57159 Sep 25 '14 at 15:53
  • $\begingroup$ Sorry about the definition. In this, case $T_i \subset \mathbb{R} - \mathbb{Q}$ for all $i$. $\endgroup$ – ppham27 Sep 25 '14 at 15:55
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We have a countable cover of $\mathbb{R}$ with the sets $(T_n)$ and $\{r\}$, $r\in \mathbb{Q}$. Since $\mathbb{R}$ is of Baire second category, the closure of one of these sets will have nonvoid interior.

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  • $\begingroup$ $\mathbb{R}$ is a complete metric space, hence of second category ( Baire's theorem). $\endgroup$ – Orest Bucicovschi Sep 25 '14 at 15:52
  • $\begingroup$ Ahh, this works. Thanks. $\endgroup$ – ppham27 Sep 25 '14 at 16:00
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    $\begingroup$ No worries. In fact, if $(A_n)_n$ is a cover of $\mathbb{R}$ then also $(\overset{\circ}{\bar A_n})_n$ is a cover of $\mathbb{R}$ ( since $\mathbb{R}$ is complete and has not isolated points). Hence any bounded interval $I$ will be contained $T_n$ for $n$ large enough ($n\ge n_I$) . $\endgroup$ – Orest Bucicovschi Sep 25 '14 at 19:21
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    $\begingroup$ Oh, it should be $I \subset \bar T_n$ for $n \ge n_I$ $\endgroup$ – Orest Bucicovschi Sep 25 '14 at 20:26

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