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Math Help - Baire Category and Residual Sets

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    Baire Category and Residual Sets

    I posted this problem some time ago, and it didn't elicit any feedback. So, in compliance with MHF Rules, I am now reposing the problem with some additional info. The result is bound to be long-winded; apologies.

    Recall the topological notion of a nowhere dense set. There are a number of equivalent definitions of this, but I shall be using the following: "A set S in \mathbb{R} is said to be nowhere dense provided that its complement S^\text{C} contains a dense, open set." If a set in \mathbb{R} can be expressed as a union of countably many nowhere dense sets, it is said to be of first (Baire) category, and if it cannot be so expressed, it is said to be of second (Baire) category. The complement of a set of first category is often called residual. It is easy to show that, in \mathbb{R}, any residual set is of second category.

    I need to prove the following:

    Let A and B be any two residual sets in \mathbb{R}. For fixed real numbers x and y, the sets A_x=\{x-a : a\in A\} and B_y = \{y/b : b\in B, b\neq 0\} are both residual in \mathbb{R}; that is, they are both complements of sets of first category.

    I can see how this is true intuitively, but I have never been able to find a rigorous proof. Any guidance anyone can provide would be of great help.
    Last edited by AlephZero; August 3rd 2009 at 03:23 PM.
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    For a fixed real number x, the map f:\mathbb{R}\to\mathbb{R} defined by f(t) = x-t is a continuous bijection. It follows that if S is dense and open then so is S_x = \{x-s:s\in S\}. In words, if S is dense and open then so is the reflected translate of S by x. The same is therefore true for the complement of S. So reflected translates of nowhere dense sets are nowhere dense. The same is therefore also true for countable unions of such sets, so reflected translates of first category sets are of first category. Taking complements again, you see that reflected translates of residual sets are residual.

    Similarly, the map g:\mathbb{R}\setminus\{0\}\to\mathbb{R}\setminus\{  0\} defined by f(t) = y/t is a continuous bijection. So you can use the same argument as for translates to show that if B is residual then so is B_y.
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    Quote Originally Posted by Opalg View Post
    For a fixed real number x, the map f:\mathbb{R}\to\mathbb{R} defined by f(t) = x-t is a continuous bijection. It follows that if S is dense and open then so is S_x = \{x-s:s\in S\}. In words, if S is dense and open then so is the reflected translate of S by x.
    Excellent! This is the part I felt was intuitively clear, but couldn't prove. Could I trouble you for a proof of this assertion? I'm sure it's probably trivial, but I am lousy at topology...
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    Quote Originally Posted by Opalg View Post
    For a fixed real number x, the map f:\mathbb{R}\to\mathbb{R} defined by f(t) = x-t is a continuous bijection. It follows that if S is dense and open then so is S_x = \{x-s:s\in S\}. In words, if S is dense and open then so is the reflected translate of S by x.
    Quote Originally Posted by AlephZero View Post
    Could I trouble you for a proof of this assertion?
    Density: For a\in\mathbb{R}, we want to show that a is in the closure of S_x. We know that S is dense, so x-a is in the closure of S. Therefore there is a sequence (s_n) in S with \lim_{n\to\infty}s_n = x-a. Therefore \lim_{n\to\infty}(x-s_n) = a. Thus a is the limit of a sequence of points in S_x and thus is in the closure of S_x.

    Openness: Given  x-s\in S_x (with s\in S), there exists \delta>0 such that for t\in\mathbb{R},\;|t-s|<\delta\;\Rightarrow\;t\in S (because S is open). Then for u\in\mathbb{R},\;|u-(x-s)|<\delta \;\Rightarrow\; |(x-u)-s|<\delta \;\Rightarrow\; x-u\in S\;\Rightarrow\; u\in S_x.
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    Lovely. Thanks very much for that wonderfully clear explanation.

    Cheers!
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