Can we judge closedness of vitali set? 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.....
 A: There is indeed a concrete proof, which avoids measure (although it does use the Baire category theorem).

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.

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.
A: Here is a more "high-level" proof.
If $\lambda^*$ is the Lebesgue outer-measure, the set of all $\lambda^*$-measurable subsets of $\mathbb{R}$ is a $\sigma$-algebra $\mathfrak{M}$ which contains the Borel $\sigma$-algebra $\mathcal{
B}$ generated by the Euclidean topology $\mathcal{T}_e$ on $\mathbb{R}$. In other words we have,
$$\mathcal{T}_e\subset \mathcal{B}\subset\mathfrak{M}.$$
Since $V$ is not $\lambda^*$-measurable, then $V\not\in\mathfrak{M}$ which means $V$ is not open.
Also $V^c\not\in\mathfrak{M}\;$(otherwise $V\in \frak{M}$, a contradictoin), so $V^c$ is not open and hence $V$ is not closed.
