Can the distance between two points equals zero Can the distance between two point on a plane be zero?
I just assumed yes but I have heard the argument no because if the points are in the same location then they are the same point and thus you are not measuring the distance between two points any more.  Others say you cannot have two points occupy the same location.  (Or can two points occupy the same location because they are dimensionless.)  Or can we say the distance is zero by some limiting argument such as point A and point B get closer and closer together the distance between goes to zero. So maybe it is never actually be zero but the limit is zero.
 A: It may seem semantic, but there is a difference to be drawn between "the distance between two points" and "the distance between two distinct points".
When a metric, or distance function, is defined on a metric space $M$, it is generally defined for all ordered pairs $(a,b)\in M\times M$, for example as a function $d:M\times M \to \mathbb R$ (other suitable ordered rings and fields are available). And it is then part of the definition that for all $x\in M$ we have $d(x,x)=0$. Also we insist that $d(x,y)=d(y,x)$, which means that we can easily lose sight of the fact that the distance applies to ordered pairs. But this all means that we can apply the distance function to the ordered pair $(x,x)$ just as to any other ordered pair.
A: In a metric space, by definition, no.  One of the axioms of metric spaces (sometimes called the "identity of indiscernibles") says that the distance between any two distinct points must be strictly greater than zero.
(Of course, if you don't require the points to be distinct, then it's perfectly possible for, say, the distance from point $A$ to point $B$ to be zero; in a metric space, this happens if and only if $A = B$.)
However, it's sometimes useful to study spaces in which two distinct points are allowed to have a zero distance.  Such spaces, if they satisfy all the other axioms of a metric space, are called pseudometric spaces.
However, it turns out that the theory of pseudometric spaces is not significantly different from that of ordinary metric spaces.  In particular, one of the other axioms of (pseudo)metric spaces, the triangle inequality, says that, for any three points $A$, $B$ and $C$, the distance from $A$ to $C$ cannot be greater than the distance from $A$ to $B$ plus the distance from $B$ to $C$.  Applied to a pseudometric space, this implies that, if the distance between points $A$ and $B$ is zero, then their distances from any other point must be equal.  Thus, any such points are indeed indistinguishable, as far as the metric is concerned.
This means that, given any pseudometric space, we can form a new space from it by identifying any points that have a zero distance from each other, and treating them as if they were the same point.  The resulting quotient space is then a full metric space, and its metric contains essentially all information about the original pseudometric, except for the number of indistinguishable points that were identified to produce each point in the new space.
A: Yes...but only if they are the same point. If you mean "can the distance between two distinct points be zero," the answer is "no".
The square of the distance between $(a, b)$ and $(p,q)$ is 
$$
(p-a)^2 + (q-b)^2
$$
Now if the distance were zero, its square would be too, so that expression would be zero. That means that $(p-a)^2 = -(q-b)^2$. But each of the squares is nonnegative, and if one were positive, the other would have to be negative. That's a contradiction, so each of the squares is zero. 
Since $(p-a)^2 = 0$, $p$ must equal $a$, and similarly for $q$ and $b$. 
A: In any norm:
$\left \| x - y \right \| = 0 \quad \Leftrightarrow  \quad x - y = 0 \quad \Leftrightarrow \quad x = y$
The physical distance is associated with the $2$-norm, which satisfies this axiom and therefore distance $0$ means same point.
