Can a set be compact if its boundary is composed by points with measure 0? I was wondering... An interval $[a,b] \in \mathbb{R}$ is closed and bounded, so it is compact. But let's appreciate that its boundary is finite and composed by two disjoint points. For me it is very clear that the set $O$ (upper case o) defined as:
$$
O = \{ (x,y) \in \mathbb{R}^2 | \sqrt{x^2 + y^2} < 1 \}
$$
is open. Also, it is clear that the following set $C$ defined as:
$$
C = \{ (x,y) \in \mathbb{R}^2 | \sqrt{x^2 + y^2} \leq 1 \}
$$
is close. Alright what about the following set:
$$
X = O \; \cup \; (\partial O \; \cap \; \mathbb{Q})
$$
Which is the full circle, but its boundary is composed only by rational points (the symbol $\partial$ gives the boundary of the set), that we know that is dense in the unit circumference. Is it open because the measure of the boundary matters and the fact that the measure of the rationals is 0 makes this set open? Is it closed because as long as you have a boundary with a dense boundary it doesn't matter the measure? Is it something in between that I don't know? The only truth that I know is the following:
$$
O \subset X \subset C
$$
Many thanks in advance!!
 A: $X$ is neither open nor closed. Let me explain why:
Open: The point $(0,1)\in X$, but you cannot find a Ball with small radius $\varepsilon>0$, such that $B_{\varepsilon}((0,1))\subset X$, because it would always contain the point $(0,1+\frac{1}{2}\varepsilon)$, which is not in $X$.
Closed: The point $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})\notin X$, because it consists of irrational values. But you can approach this point by a sequence in $O$, for example 
$(\frac{1}{\sqrt{2}}-\frac{1}{n},\frac{1}{\sqrt{2}}-\frac{1}{n})$, hence the closure of $X$ is not $X$ itself.
The message is, that the measure of the boundary has in general no impact on the set being compact. 
A: $X$ is not open, since each $ (x_0,y_0) \in \partial O \cap \mathbb Q$ is a member of $O$, but not an interior point of $O$.
$X$ is not closed, since $ \overline{X}=C \ne X.$
We have $O \subset X \subset C$, but $O \ne  X \ne C.$
A: The meaure has nothing to do with openess/closeness of the set. These are purely topological concepts. The fact that you chose a dense subset in the boundary doesn't change anything.
Your set is neither open or closed. Comapre with one dimensional example of intervals:
$$ (a,b) \subset  (a,b] \subset [a,b]$$
The first one is open, the third one is closed, the second one is neither.
