Baire Category and Residual Sets

**AlephZero** · August 3rd 2009, 08:47 AM

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.

**Opalg** · August 3rd 2009, 01:15 PM

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 $B_y$ .

**AlephZero** · August 3rd 2009, 02:21 PM

Originally Posted by Opalg

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...

**Opalg** · August 4th 2009, 08:30 AM

Originally Posted by Opalg

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.

Originally Posted by AlephZero

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 $u\in\mathbb{R},\;|u-(x-s)|<\delta \;\Rightarrow\; |(x-u)-s|<\delta \;\Rightarrow\; x-u\in S\;\Rightarrow\; u\in S_x.$

**AlephZero** · August 4th 2009, 09:55 AM

Lovely. Thanks very much for that wonderfully clear explanation.

Cheers!

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