Suppose we have a cylindrical resistor, with resistance given by $R=\rho\cdot l/(\pi r^2)$
Let $d$ be the distance between two points in the interior of the resistor and let $r\gg d\gg l$.
Ie. it is approximately a 2D-surface (a rather thin disk).
What is the resistance between these two points?
Let $r,l\gg d$, (ie a 3D volume), is the resistance $0$ ?
Clarification:
A voltage difference is applied between two points a distance $d$ apart, inside a material with resistivity $\rho$, and the current is measured, the proportionality constant $V/I$ is called $R$. The material is a cylinder of height $l$ and radius $r$, and the two points are situated close to the center, we can write $R$ as a function of $l$, $r$ and $d$, $R(l,r,d)$, for small $d$.
The questions are then:
What is $$ \lim_{r \rightarrow \infty} \lim_{l \rightarrow \infty} R(l,r,d) $$
What is $$ \lim_{r \rightarrow \infty} \lim_{l \rightarrow 0} R(l,r,d) $$
Best Answer
Potential for 2D problem
Let's start with a 2D disk and try to solve the general problem for infinitesimally flat disk. I will change notations a bit -- the surface resistance will be $\sigma$ and the radius of the disk will be $a$.
Starting with basic electrodynamics:
$\vec{j} = -\sigma\frac{\partial u}{\partial \vec{r}},\, div\vec{j}=0\,\Rightarrow\,\Delta u = 0$ with the boundary condition: $\vec{n}\cdot\vec{j} = 0 \Rightarrow \vec{n}\frac{\partial u}{\partial \vec{r}} = 0$
Let's first consider the current $I$ flowing into the surface in the centre and uniformly flowing away from the edges. solution for potential is well known:
$U(r,\phi) = -\frac{I}{2\pi\sigma}\, \ln r$
I use the conformal map $z\to a\frac{z-s}{a^2-s*z}$ to "shift the centre" into the point $s=x_{source}+iy_{source}$. The potential is then:
$U(r,\phi) = -\frac{I}{2\pi\sigma}\, \ln\left|a\frac{re^{i\phi}-s}{a^2-s^*re^{i\phi}}\right|$
Now I substract the similar potential, with different parameter $d=x_{drain}+iy_{drain}$ to compensate the outgoing flow. Obtaining:
$U(r,\phi) = -\frac{I}{2\pi\sigma}\, \ln\left|\frac{re^{i\phi}-s}{re^{i\phi}-d} \cdot\frac{a^2-d^*re^{i\phi}}{a^2-s^*re^{i\phi}}\right|$ or $U(z) = -\frac{I}{2\pi\sigma}\, \ln\left|\frac{z-s}{z-d} \cdot\frac{a^2-d^*z}{a^2-s^*z}\right|$
This is the harmonic function, satisfying the boundary conditions. You can play here with it.
Interpretation of the solution
The potential is divergent in points $s$ and $d$. This happens because the resistance is strongly dependend on the microscopic details of the problem. Indeed -- as you get closer to the source -- all your current have to pass through smaller and smaller amount of conductor. And in the limit of infinitely small source you get infinite resistance.
Formulation issue
I admit that while solving I first fixed the current and then found the potential, while you formulated the problem differently -- "set the potential here and there and find the current". But let us use logic:
At finite voltage you'll get zero current or, equivalently, infinite resistance.
What happens in 3D case?
Same thing. Just consider single pointlike source -- and the potential $U\sim\frac{1}{r}$ is divergent. Don't need to go into further details.
"Cutoffs"
In order to move on I introduce the "cutoffs" -- new small (real) quantities $\epsilon_{s,d}$ which denoting "sizes" of the source and the drain. Using them I obtain the voltage:
$U(d+\epsilon_d)-U(s+\epsilon_s)=\frac{I}{2\pi\sigma}\left[\ln\frac{\epsilon_s}{|s-d|}+\ln\frac{\epsilon_d}{|s-d|} +\ln\left|\frac{a^2-s^*d}{a^2-|s|^2}\cdot\frac{a^2-sd^*}{a^2-|d|^2}\right|\right]$
Scales
Putting together everything above. One can say that in the problem there are four (or, even five) scales:
Since you are talking about "points" -- then first we have to take $\epsilon_{s,d}\to0$, right? But if $\epsilon_{s,d}$ is much smaller that any other scale then they introduce divergent contribution into the resistance. And any other detail of the problem becomes irrelevant.
Therefore, the answer to your question is: The resistance between two points is infinite, whatever the geometry of the problem is.