If a pseudonorm function $N$ is continuous in a given topology, does the pseudometric topology formed by $d(x,y)=N(x-y)$ coincide with first topology?

Question

Define $p_n\#= p_n p_{n-1} \cdots p_1$ for $n = 0$ to be $1$, then we have a function:

$$ N : \Bbb{Z} \to \Bbb{Z}, \\ N(x) = \left |-1/2 + \sum_{d\ \mid\ p_n\#} (-1)^{\omega d} \sum_{r^2 = 1 \mod d} \left\lfloor \frac{x - r}{d}\right\rfloor \right |\\ \text{ with } n = n(x) = \pi(\sqrt{|x| + 1}) $$

where $\pi$ is the usual prime counter.

Here is the output of SymPy code that computes $N(x)$ and $N(-x)$ side-by-side:

N(-0) = 0.5;  N(0) = 0.5
N(-1) = 1.5;  N(1) = 0.5
N(-2) = 2.5;  N(2) = 1.5
N(-3) = 1.5;  N(3) = 1.5
N(-4) = 1.5;  N(4) = 2.5
N(-5) = 2.5;  N(5) = 2.5
N(-6) = 2.5;  N(6) = 3.5
N(-7) = 3.5;  N(7) = 3.5
N(-8) = 1.5;  N(8) = 1.5
N(-9) = 1.5;  N(9) = 1.5
N(-10) = 1.5;  N(10) = 1.5
N(-11) = 1.5;  N(11) = 1.5
N(-12) = 1.5;  N(12) = 2.5
N(-13) = 2.5;  N(13) = 2.5
N(-14) = 2.5;  N(14) = 2.5
N(-15) = 2.5;  N(15) = 2.5
N(-16) = 2.5;  N(16) = 2.5
N(-17) = 2.5;  N(17) = 2.5
N(-18) = 2.5;  N(18) = 3.5
N(-19) = 3.5;  N(19) = 3.5
N(-20) = 3.5;  N(20) = 3.5
N(-21) = 3.5;  N(21) = 3.5
N(-22) = 3.5;  N(22) = 3.5
N(-23) = 3.5;  N(23) = 3.5
N(-24) = 2.5;  N(24) = 2.5
N(-25) = 2.5;  N(25) = 2.5
N(-26) = 2.5;  N(26) = 2.5
N(-27) = 2.5;  N(27) = 2.5
N(-28) = 2.5;  N(28) = 2.5
N(-29) = 2.5;  N(29) = 2.5
N(-30) = 2.5;  N(30) = 3.5
N(-31) = 3.5;  N(31) = 3.5
N(-32) = 3.5;  N(32) = 3.5
N(-33) = 3.5;  N(33) = 3.5
N(-34) = 3.5;  N(34) = 3.5
N(-35) = 3.5;  N(35) = 3.5
N(-36) = 3.5;  N(36) = 3.5
N(-37) = 3.5;  N(37) = 3.5
N(-38) = 3.5;  N(38) = 3.5
N(-39) = 3.5;  N(39) = 3.5
N(-40) = 3.5;  N(40) = 3.5
N(-41) = 3.5;  N(41) = 3.5
N(-42) = 3.5;  N(42) = 4.5
N(-43) = 4.5;  N(43) = 4.5
N(-44) = 4.5;  N(44) = 4.5
N(-45) = 4.5;  N(45) = 4.5
N(-46) = 4.5;  N(46) = 4.5
N(-47) = 4.5;  N(47) = 4.5
N(-48) = 4.5;  N(48) = 4.5
N(-49) = 4.5;  N(49) = 4.5
N(-50) = 4.5;  N(50) = 4.5
N(-51) = 4.5;  N(51) = 4.5
N(-52) = 4.5;  N(52) = 4.5
N(-53) = 4.5;  N(53) = 4.5
N(-54) = 4.5;  N(54) = 4.5
N(-55) = 4.5;  N(55) = 4.5
N(-56) = 4.5;  N(56) = 4.5
N(-57) = 4.5;  N(57) = 4.5
N(-58) = 4.5;  N(58) = 4.5
N(-59) = 4.5;  N(59) = 4.5
N(-60) = 4.5;  N(60) = 5.5
N(-61) = 5.5;  N(61) = 5.5
N(-62) = 5.5;  N(62) = 5.5
N(-63) = 5.5;  N(63) = 5.5
N(-64) = 5.5;  N(64) = 5.5
N(-65) = 5.5;  N(65) = 5.5
N(-66) = 5.5;  N(66) = 5.5
N(-67) = 5.5;  N(67) = 5.5
N(-68) = 5.5;  N(68) = 5.5
N(-69) = 5.5;  N(69) = 5.5
N(-70) = 5.5;  N(70) = 5.5
N(-71) = 5.5;  N(71) = 5.5
N(-72) = 5.5;  N(72) = 6.5
N(-73) = 6.5;  N(73) = 6.5
N(-74) = 6.5;  N(74) = 6.5
N(-75) = 6.5;  N(75) = 6.5
N(-76) = 6.5;  N(76) = 6.5
N(-77) = 6.5;  N(77) = 6.5
N(-78) = 6.5;  N(78) = 6.5
N(-79) = 6.5;  N(79) = 6.5
N(-80) = 6.5;  N(80) = 6.5
N(-81) = 6.5;  N(81) = 6.5
N(-82) = 6.5;  N(82) = 6.5
N(-83) = 6.5;  N(83) = 6.5
N(-84) = 6.5;  N(84) = 6.5
N(-85) = 6.5;  N(85) = 6.5
N(-86) = 6.5;  N(86) = 6.5
N(-87) = 6.5;  N(87) = 6.5
N(-88) = 6.5;  N(88) = 6.5
N(-89) = 6.5;  N(89) = 6.5
N(-90) = 6.5;  N(90) = 6.5
N(-91) = 6.5;  N(91) = 6.5
N(-92) = 6.5;  N(92) = 6.5
N(-93) = 6.5;  N(93) = 6.5
N(-94) = 6.5;  N(94) = 6.5
N(-95) = 6.5;  N(95) = 6.5
N(-96) = 6.5;  N(96) = 6.5
N(-97) = 6.5;  N(97) = 6.5
N(-98) = 6.5;  N(98) = 6.5
N(-99) = 6.5;  N(99) = 6.5
N(-100) = 6.5;  N(100) = 6.5
N(-101) = 6.5;  N(101) = 6.5
N(-102) = 6.5;  N(102) = 7.5

As you can see, it doesn't quite satisfy certain types of norms' axiom $N(-x) = N(x)$. Since at the twin prime averages $x$ we have that $N(-x) +1 = N(x)$. However, under the assumption that the twin primes end, we must have that for sufficiently large $x \geq 0$, $x \in \Bbb{Z}$ that $N(x) = N(-x)$.

Remark. We may also take $N : \Bbb{R} \to \Bbb{Z}$ to be the function (domain is $\Bbb{R}$) and the above should still hold. So that gives us the freedom to work with either an abelian group $\Bbb{Z}$ or a vector space $\Bbb{R}$.

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Now also under the assumption that the twin primes end, we must have that for sufficiently large $x, y \geq 0$ that $N(x + y) \leq N(x) + N(y)$. This is because under that assumption, we have that $N(x)$ eventually equals $0$ and so of course $0 = N(x + y)\leq N(x) + N(y)$ as $N(x) \geq 0$ in general.

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Question. If we can show that $N(x)$ forms a pseudonorm for sufficiently large inputs and we define under similar "sufficiently large inputs" a pseudometric via $d(x,y) = N(x - y)$. And, we know that $N(x)$ is continuous in a given topology (it's likely continuous in the Furstenberg topology), then can we conclude that the pseudometric topology induced by $N$ is indeed the same as the Furstenberg topology?

Easier Related Question. Forget about "sufficiently large inputs" in the previous question. If $N$ were a usual pseudonorm, and we have proven that $N(x)$ is a continuous function $\Bbb{Z} \to \Bbb{Z}$ (e.g. under Furstenberg's topology), then can we conclude that the pseudometric topology induced by $d(x,y) = N(x-y)$ coincides with the given topology on $\Bbb{Z}$?

Answer 1

No. In Furstenberg's topology $\phi$ on $\Bbb{Z}$ in particular. The open balls formed by the norm would look like:

$$ B_r(y) = \{ x \in \Bbb{Z} : N(y-x) \lt r\} = N(y-\cdot)^{-1} \{1, \dots, \lfloor r \rfloor\},\\ \forall \ r \gt 0 $$

Finite sets are closed in any metric space and the Furstenberg topology is known to be metrizeable.

Therefore, the open balls in the norm-induced topology are all closed sets in the Furstenberg topology, because $(\Bbb{Z}, \phi)$ is a topological ring, making $f(x) = y - x$ continuous and thus the composition $N(f(x)) = N(y - x)$ is also a continuous function.

This is kind of interesting, but I was hoping for coinciding topologies -_-