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Show (i) the translate of an $F_{\sigma}$ set is also $F_{\sigma}$, (ii) the translate of a $G_{\delta}$ set of a $G_{\delta}$, and (iii) the translate of a set of measuure zero also has measure zero.

I'm a little confused, because...

$F_{\sigma}$ is of the form $\cup^{\infty}_{n=1} [a_n, a_{n+1}]$, right? So if we translate this by y, we get $\cup^{\infty}_{n=1} [a_n+y, a_{n+1}+y]$, right? Isn't this already in $F_{\sigma}$ $ form? I'm not really sure what the question wants me to do...

Thanks in advance

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Note that the Cantor set is a closed (and hence F$_\sigma$) set, but is not the countable union of closed intervals. F$_\sigma$ sets are countable unions of closed sets, but these closed sets need not be intervals. Though the basic idea behind your approach is correct: translates of open sets are open; translates of closed sets are closed; etc. –  Arthur Fischer Feb 6 '13 at 10:09
    
@ArthurFischer Thanks, but I didn't read about Cantor sets yet...Also, what do you mean by "the basic idea behind your approach is correct"? I didn't prove anything, because I'm confused about what the question is asking for.. –  user58289 Feb 6 '13 at 10:30
    
I'll include some details in an answer below. –  Arthur Fischer Feb 6 '13 at 10:38

1 Answer 1

up vote 2 down vote accepted

It appears that you have the right idea without knowing it. Basically, you are asked to show the following: Suppose that $A \subseteq \mathbb{R}$ is a set with some property, show that all translates $A + y$ of this set have the same property. The details become different as we go from property to property, but the idea behind the proofs are the same: we will take witnesses to the defining characteristic of the particular property in question and then translate them. We then only have to show that these translates then witness that $A + y$ has that same property.

Consider the property "is F$_\sigma$." Now, if $A \subseteq \mathbb{R}$ is F$_\sigma$ this means that there is a countable family $\{ E_n : n \in \mathbb{N} \}$ of closed sets such that $A = \bigcup_n E_n$. To show that $A + y$ is also F$_\sigma$ we must find another countable family $\{ F_n : n \in \mathbb{N} \}$ of closed subsets of $\mathbb{R}$ such that $A + y = \bigcup_n F_n$. Well, one thing we know is that $$A + y = ( {\textstyle \bigcup_n} E_n ) + y = {\textstyle \bigcup_n} ( E_n + y ).$$ Could these sets all be closed? Yes: If $x \notin E_n + y$, this means that $x - y \notin E_n$ and since $E_n$ is closed there is a $\epsilon > 0$ such that $( ( x - y ) - \epsilon , ( x - y ) + \epsilon ) \cap E_n = \emptyset$, and we can then show that $( x - \epsilon , x + \epsilon ) \cap ( E_n + y ) = \emptyset$; this shows that $E_n + y$ is closed.

The details for "is G$_\delta$" are almost entirely the same. (Translates of open sets will be open sets.)

For "has measure zero" note that for $A$ to have measure zero then for each $\epsilon > 0$ there is a family $\{ I_n : n \in \mathbb{N} \}$ of open intervals in $\mathbb{R}$ such that $A \subseteq \bigcup_n I_n$, and $\sum_n \mathrm{length} ( I_n ) < \epsilon$. (Translates of open intervals will be open intervals of the same length.)

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Thanks, but I am just a little confused about how we get $(x-\epsilon, x+\epsilon) \cap (E_n+y) = \emptyset$ from $((x-y)-\epsilon, (x-y)+\epsilon) \cap E_n = \emptyset$. –  user58289 Feb 9 '13 at 9:30
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@Artus: If $z \in (x - \epsilon, x + \epsilon) \cap ( E_n + y )$ then consider $z - y$. Since $z \in E_n + y$ it follows that $z - y \in E_n$. And since $x - \epsilon < z < x + \epsilon$ it follows that $( x - y ) - \epsilon < z - y < ( x - y ) + \epsilon$. –  Arthur Fischer Feb 9 '13 at 9:36

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