Curve of zeta(0.5 + i t) : "Dense" on complex plane?

[Swamp Thing]

This is a discussion on MathOverflow where a conjecture is discussed that the curve of ##\zeta(0.5+it)## is "dense" on the complex plane.
https://mathoverflow.net/questions/...of-riemann-zeta-function-on-the-critical-line

From a couple of sources, e.g. www.reddit.com/r/math/comments/8c9uh7/dense_sets_in_the_complex_plane/
I learned that "dense" means this: No matter how small a circle you draw and no matter where it's centered, the dense set has at least one point inside the circle.

From the following argument, it appears that the set of zetas is not dense. What is the flaw in it (if any)?

Let's look at the curve from t=0 up to, say, the 100th zero. We see that it divides the plane into a lot of areas that look like diamonds or like stretched-out diamonds, or crescent like slivers. Let's pick a point in the middle of one of these areas and draw a circle that is entirely within it. Now let's extend our curve from the 100th zero to the 101st zero. If the extended curve doesn't pass through our circle, we're still good. But if the curve now cuts our circle, it means that our chosen diamond subdivision has been further subdivided. So we just dodge away from the curve by choosing a new point within one of the new smaller subdivisions. We choose a circle that lies within that, rinse and repeat.

Thus if we extend the curve through any number of zeros, we can keep dodging into the smaller and smaller subdivisions (when required). If we take this to infinity, we get a point that is the sum of our original position, plus a potentially infinite convergent series of evasive moves. This complex number is, in one sense, not on the curve for t -> infinity.

I'm not sure if the above is correct. Maybe I've fallen into a sort of Zeno's paradox, and have wrongly interpreted the definition of dense. Maybe the answer depends on the relative rate at which t has to grow, versus the rate at which our circle shrinks?

Any clarifications greatly appreciated.

 

[Infrared]

You seem to be claiming that a continuous curve can never be dense in the plane (since you don't use any fact about the zeta function other than it being continuous). Here is a counterexample: Let ##q_1,q_2,\ldots## be an enumeration of rational points in the plane (i.e. both coordinates are rational). Then take a piecewise-linear curve that starts at ##q_1##, goes to ##q_2##, etc. Then every rational point lies on the curve, so the curve is dense in the plane.

Edit: You can in fact have curves whose image is the entire plane (not just dense). Search "space-filling curves" for explicit constructions.

 

[Swamp Thing]

I think I get it now -- to prove that the curve is not dense, we would have to prove that there is at least one area that never gets further subdivided, however high we take t?