$\Bbb{Q}$ is dense in $\Bbb{R}$. Let $\mathrm{B}$ a finite set. $\Bbb{Q}$\ $\mathrm{B}$ is still dense in $\Bbb{R}$? And if $\mathrm{B}$ is infinite? The concept behind density is:
"A subset B $\subset$ X is dense in X when $\overline{B}=X$"
I`m thinking, if the subset $B$ is just one point (then, finite), so $\Bbb{Q}$\ $\mathrm{B}$ continue dense in $\Bbb{R}$. My argument is that we take the real line and observe the $\Bbb{Q}$ numbers and take off one (Real or Rational), the set will continue being a subset of $\Bbb{R}$. In the same way, $\Bbb{R}$ contain $\Bbb{Q}$ without the same point.
Now... as you can see, guys, I'm completely lost. I'm not so sure what i'm talking about. At least, i created an argument (half-assed) to finite, but to infinite, i feel myself nailed to the floor. 
Please help!
 A: Here I would use the fact that a set $D$ is dense in a space $X$ if and only if $U\cap D\ne\varnothing$ whenever $U$ is a non-empty open set in $X$. Let $B$ be a finite subset of $\Bbb Q$, and let $D=\Bbb Q\setminus B$; we want to show that $D$ is still dense in $\Bbb R$. Let $U$ be any non-empty open set in $\Bbb R$. 


*

*Explain why $U$ is infinite and hence why $U\setminus B$ is non-empty.  

*Explain why $U\setminus B$ is open.  

*Use the previous point and the fact that $\Bbb Q$ is dense in $\Bbb R$ to show that $D\cap U\ne\varnothing$.


If $B$ is infinite, anything can happen: $\Bbb Q\setminus\Bbb Z$ is still dense in $\Bbb R$, while $\Bbb Q\setminus\Bbb Q=\varnothing$ very obviously is not! (It’s a good little exercise to verify that $\Bbb Q\setminus\Bbb Z$ is dense in $\Bbb R$.)
A: Ugh.  My first answer was terrible.  Here's a redo:
"The concept behind density is: "A subset B ⊂ X is dense in X when $\overline B=X".
Although this is a valid and complete definition of "dense", in my opinion, it's not a very good one for concept or intuition unless one has an intuitive sense of precise what $\overline B$ conceptually is.
Speaking for myself, I don't.  Or maybe I do now after a lot of experience.
I mean, sure, I get that the closure of $(1,5)$ is $[1,5]$ because we are adding in the endpoints.  But how is $\overline{\mathbb Q} = \mathbb R$-- $\mathbb Q$ doesn't have any endpoints?  Okay, I understand maybe we are "filling in all the blanks" but then why is $\overline{\{1/n|n \in \mathbb N\}} = \{1/n|n \in \mathbb N\} \cup \{0\}$? $0$ is neither an "endpoint" nor is it "filling in" any "gaps".
In my opinion the most fundamental concept to understand about any analytic set concept is "limit point". A limit point of a set is a point that is arbitrarily close to points of the set.  Or more accurately, a limit point of a set is a point so that any arbitrary distance from the point will contain points of the set (different than itself). 
$1$ is a limit point of $(1,5)$ because for any distance $\epsilon > 0$, I can find $y \in (1,5)$ so that $1 < y < 1 + \epsilon$.  $.95$ is not a limit point of $(1,5)$ because I can't find any points $y \in (1,5)$ so that $.95 < y < .95 + .04$ for $\epsilon = .04$.
Any real number, $x$ is a limit point of $\mathbb Q$ because for any $\epsilon > 0$ I can find $q \in \mathbb Q$ so that $x -\epsilon < y < x+\epsilon$.
$0$ is a limit point of $\{1/n\}$ because for any $\epsilon > 0$ I can find a $1/n$ so that $0 < 1/n < \epsilon$.  $0$ is the only limit point of $\{1/n\}$ by the way, because for any $x \ne 0$: i) if $x < 0$ then there is no 1/n so that $x < 1/n < x+\epsilon = 0$ for $\epsilon +|x|$.  If $x > 1$ then for $epsilon = x - 1$ there are no $1/n$ so that $1 = x - \epsilon < 1/n < x$.  If $x = 1/k$ then for $\epsilon = 1/k - 1/(k+1)$ there is not $1/n$ (other than $1/k$) so that $1/k-1 = x - \epsilon < 1/n < x + 1/(k+1)$.
The formal definition of a limit point is: $x$ is a limit point of $S$ if for every real $\epsilon > 0$ there exists a $y \in S; y \ne x$ so that $d(x,y) < \epsilon$.  (In the reals with the euclidean metric $d(x,y) =|x-y|$.)  Or alternatively $x$ is a limit point of $S$ if every open neighborhood, $B(x,\epsilon) = \{y \in \text{the universal metric space, X}| d(x,y) < \epsilon\}$ will contain a point $w \in S; w \ne x$.  (Those two definitions are essentially the same.)
Basically a limit point, $x$ of $S$, is any point that for an arbitrary distance there are points of $S$ that are closer than that distance to $x$.
Note: a limit point may or may not be an element of the set.  And an element of the set may or may not be a limit point.
Also note: if the set $S$ has a limit point.  Then $S$ must have infinite elements. Why, because if $S$ has a finite number of points, then for any point $x$ we can find an $s \in S; s$ so that $d(x,s) = |x-s|$ is the smallest value.  (If $S$ is infinite there might not be a smallest value but if $S$ is finite there must be.)  If we let $\epsilon = d(x,s)$ then there are not point $t\in S$ so that $d(x,t) < \epsilon$.  So $x$ is not a limit point.  So no finite sets have limit points.
Okay... now that we have an intuitive idea of limit points, we can develop an intuitive idea of what a "closed set" means.   
Remember I said a limit point of $S$ might or might not an element of $S$.  $0 \not \in \{1/n\}$.  $x \text{ irrational } \not \in \mathbb Q$, $1,5 \not \in(1,5)$.
A set, $S$,  is "closed" if all the limit points of $S$ are members of $S$.  So none of those sets are closed.  However $\overline {S} = S \cup \{\text{the limit points of } S\}$ would be a closed set.  The "closure" of a set is simply the set plus all it's limit points.
$\overline{(1,5)} = [1,5]$ and $\overline {\mathbb Q} = \mathbb R$ and $\overline {\{1/n\}} = \{1/n\} \cup \{0\}$.
Okay... so now let's talk about density.
A set $B$ is dense in $X$ if $\overline B = B\cup \text{limit points of } B = X$.  Or in other words. $B$ is dense in $X$ if every point in $X$ is either a point of $B$ or a limit point of $B$.
So $\mathbb Q$ is dense in $\mathbb R$ because every point of $\mathbb R$ is either rational or irrational, and every irrational number is a limit point of $\mathbb Q$.
So what if we remove a finite number of points from $\mathbb Q$.  Would it still be dense in $\mathbb R$?
Well suppose $x \in \mathbb R$.  $x$ is a limit point of $\mathbb Q$ but is it a limit point of $\mathbb Q$ with a finite number of points removed?  Well, let $S$ equal the finite set of remove points. There must be some point $s \in S; s\ne x$ that is the closest to $x$.  Let $\epsilon = d(x,s) = |x - s| > 0$. $x$ is a limit point of $\mathbb Q$ so there is a $q \in \mathbb Q$ such that $d(x,q) < \epsilon$ so $q \not \in S$ as $d(x,q)$ is smaller than any $d(x,s_i); s_i \in S$.  So $q$ was not one of the points removed from $\mathbb Q$.  so $q \in \mathbb Q \setminus S$.  So $x$ is a limit point of $\mathbb Q \setminus S$.
So $\mathbb Q \setminus S$ is dense is $\mathbb R$.
What if we remove an infinite set of points for $\mathbb Q$. Would the resulting set, $\mathbb Q \setminus S$ be dense in $\mathbb R$?  Well, it could be.  $\mathbb Q \setminus \{1/n\}$ is dense in $\mathbb R$.  (I'll leave it as an exercise.) But it doesn't have to be.  After all $\mathbb Q$ is infinite and $\mathbb Q \setminus \mathbb Q = \emptyset$ and $\emptyset$ is clearly not dense in $\mathbb R$.
But more generally if we remove all the points within some finite distance of a point.  That point can no longer be a limit point.  e.g.  $B= \mathbb Q \setminus (a,b)$ is not dense in $\mathbb R$ because $\frac {a+b}2$ is not a limit point.  There is no $y\in B$ such that $a= \frac{a+b}2 - \frac{b-a}2 < y < b=\frac{a+b}2 + \frac{b-a}2$. So $\frac{a+b}2$ is not a limit point of $B$.
