Can we judge closedness of vitali set?

Question

I am not able to judge whether the vitali set is closed or not( in its parent set or u can say reals)..... Since it is 'mainly ' a subset of irrationals(except one rational representative) and irrationals are not closed subset of reals....I believe(am not sure) vitali set is not closed..... But if I construct a set by taking irrational representatives in such a way that these form a sequence with 0(or any rational) as the limit point and the rational representative to be the same limit point.......then can I say that this set is closed??? M very confused.....kindly help me understand this concept..... Any help will be heartily appreciated.....

Answer 1

There is indeed a concrete proof, which avoids measure (although it does use the Baire category theorem).

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Let's talk about a Vitali set $V$ in $\mathbb{R}$, rather than in $[0, 1]$, for simplicity (so $V$ is a set of reals such that for each $r\in\mathbb{R}$, there is exactly one $s\in V$ with $s-r\in\mathbb{Q}$).

Consider the function $f_V$ associated to $V$, defined as $f_V(r)=s-r$ where $s$ is the unique element of $V$ such that $r-s\in\mathbb{Q}$. Note that we'll always have $f_V(r)\in\mathbb{Q}$, and $f_V(r)+r\in V$ for all $r$.

We claim - and will justify this below - that for some $q\in\mathbb{Q}$ and some nonempty interval $(a, b)$, the set $X=f_V^{-1}(q)\cap (a, b)$ is dense in $(a, b)$. If so, then:

• Note that for each $r\in X$, we have $r+q\in V$.

• If $V$ is closed, this means that $(a+q, b+q)\subseteq V$ (since $\{x+q: x\in X\}$ is contained in $V$ and is dense in $(a+q, b+q)$).

• But $(a+q, b+q)$ contains many distinct rationals - a contradiction.

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So how do we prove the existence of such $q, a, b$? It comes down to the following fact:

If $g:\mathbb{R}\rightarrow\mathbb{Q}$, then there is some $q\in\mathbb{Q}$ and nontrivial interval $(a, b)$ such that $g^{-1}(q)\cap (a, b)$ is dense in $(a, b)$.

And this is just the Baire category theorem: that $\mathbb{R}$ isn't the union of countably many nowhere-dense sets.