Equivalence of the Box/Ball topology

From Vectorcalcumb

So, when we say that the ball and box topology are equivalent, we mean that for any neighbhorhood in either, there is a neighbhorhood in the other which contains the original neighbhorhood, right?

We don't mean that there is a neighbhorhood in the other which contains all the same points, correct? On going to write just now, I had thought that this was clearly impossible, since you can't square the circle, but apparently you can square the circle, just not with a ruler and compass(Which, of course makes perfect sense, it's just a square with <m>\\sqrt{\\pi} r</m> length sides). However, it still seems to me that even if we have a neighbhoorhood of equal area(volume, extent) around a point, we can't possibly have an elementwise identical neighbhoorhood in the other topology, just one defined to have the same extent in the other topology. I guess that's the only kind of equivalence we need for what we want to do...