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I'm having difficult to solve this: Determine whether or not $\mathbb{R}_l$ is a Baire space.

I tried to aplly the following lemma: "X is a Baire space iff given any countable collection $\mathbb{U}_n$ of open sets in X, each of which is dense in X, their intersection $\bigcap{U}_n$ is also dense in X."

Where $\mathbb{R}_l$ is the topology of lower limit, generated by the intervals [a,b).

Could you help to solve this? Thank you.

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  • $\begingroup$ Is $\mathbb{R}_l$ an unorthodox way to denote the $l$-dimensional real space? If so, there is a famous theorem that can help, "A complete metric space is a Baire space. Also, a locally compact space is a Baire space." $\endgroup$ – Daniel Fischer Jun 26 '13 at 23:43
  • $\begingroup$ I'm not sure if $\mathbb{R}_l$ defined like above satisfies the conditions of the Baire theorem. Thank you @DanielFischer. $\endgroup$ – User43029 Jun 26 '13 at 23:50
  • $\begingroup$ Ah, that's an entirely different matter. Would have been too easy. I don't know offhand whether that's a Baire space. $\endgroup$ – Daniel Fischer Jun 26 '13 at 23:51
  • $\begingroup$ Thank anyway @DanielFischer I'll keep trying here. $\endgroup$ – User43029 Jun 26 '13 at 23:56
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    $\begingroup$ See this. $\endgroup$ – David Mitra Jun 27 '13 at 0:20
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Suppose topologies $\sigma$ and $\tau$ on a set $X$ are such that every nonempty set from $\sigma$ contains a nonempty set from $\tau$, and the other way around. This is the case for the standard and lower-limit topology on the real line. You should be able to show that, for $Y \subset X$,

  • the $\sigma$-interior of $Y$ is empty if and only if the $\tau$-interior of $Y$ is empty.
  • $Y$ is $\sigma$-nowhere dense if and only if $Y$ is $\tau$-nowhere dense.

It follows that $\sigma$ is Baire if and only if $\tau$ is Baire.

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    $\begingroup$ @Charles: No problem! Here's something else which might be fun to think about. A standard application of Baire's theorem is to showing the nowhere differentiable functions are (uniformly) dense in the continuous functions $\mathbb{R} \to \mathbb{R}$. I think continuous functions $\mathbb{R}_\ell \to \mathbb{R}$ are the same thing as "left continuous" functions $\mathbb{R} \to \mathbb{R}$. Maybe you can use your result to prove that the space of left continuous functions $\mathbb{R} \to \mathbb{R}$ has to densely contain some similarly nonintuitive set of functions. $\endgroup$ – Mike F Jul 1 '13 at 6:32
  • $\begingroup$ Ow nice @Mike! I will think about it! $\endgroup$ – User43029 Jul 2 '13 at 13:49
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Here is a more direct proof.

Let $U_n$ be open and dense in $\mathbb{R}_l$. Consider $\text{Int}(U_n)$ where the interior is taken with respect to $\mathbb{R}$. Then $\text{Int}(U_n)$ is open in $\mathbb{R}$. To show that $\text{Int}(U_n)$ is dense in $\mathbb{R}$, let $V$ be any open set of $\mathbb{R}$. Then $V$ is also open in $\mathbb{R}_l$ so $V\cap U_n$ is non-empty and open in $\mathbb{R}_l$.This means $V\cap U_n$ contains a basis element of the form $[a,b)$ (where $a<b$), so it contains an interval $(a,b)$. In the topology of $\mathbb{R}$, $(a,b)$ is open so $(a,b)\subseteq \text{Int}(U_n)\cap V$. This proves that $\text{Int}(U_n)$ is dense in $\mathbb{R}$.

Since $\mathbb{R}$ is a Baire space, $\bigcap \text{Int}(U_n)$ is dense in $\mathbb{R}$. Therefore, $\bigcap U_n$ is dense in $\mathbb{R}$. To show that $\bigcap U_n$ is dense in $\mathbb{R}_l$, it suffices to show that any given basis element $[a,b)$ intersects $\bigcap U_n$. But $[a,b)$ contains $(a,b)$ which intersects $\bigcap U_n$ since the latter is dense in $\mathbb{R}$.

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