A note about the latter article wants a version of the illustration with the image reversed. Perhaps this will do. Firefox and Safari for Mac show it nicely but I have not tested it elsewhere. Note that in the first part of the development the point moves down, instead of up, as y increases. This may have caused the discrepancy.

I think that the Z-order article is good but it should note that the classic space filling curves are continuous while the ‘Z curve’ is not.

Z: [0, 1] → [0, 1] × [0, 1]
Z: x ↦ <u(x), v(x)>

If x = ∑x_i2⁻ⁱ then
u(x) = ∑x_2i 2⁻ⁱ and
v(x) = ∑x_2i−1 2⁻ⁱ.
x_i ∊{0, 1} and i ranges from 1 to ∞.

as x approaches 1/2 from below, (x ≃ .011111111...) <u(x), v(x)> approaches <1, 1/2> but as x approaches 1/2 from above, (x ≃ .100000000...) <u(x), v(x)> approaches <0, 1/2>. The long line in the middle of the bottom picture is a give-away. Thus Z is discontinuous at 1/2 and indeed at k2⁻ⁿ for all n > 0 and 0 < k < 2ⁿ.

Likewise (Z(x+∆x) − Z(x))/∆x → <∞, ∞> when ∆x is of the form 2⁻ⁿ and thus Z is nowhere differentiable even where it is continuous, which is almost everywhere.

Z is like many other prominent space filling curves, however, in that it preserves Lebesgue measure.