I forgot to note $\left|\ell\cap[a,b]\right|=\mathfrak{c}$. I can't make more edits or I could risk being suspended.

What does "$\ell'\to \ell$" mean? Also, per one of your linked posts, there is no hyper discontinuous function on any interval $[a,b]$, and so this is probably not worth considering when building this metric.

Also, what role do the sets $X$ and $Y$ play in your definition? It seems like $f$ has to be a function from $\mathbb{R}$ to $\mathbb{R}$ in order for $\mathcal{D}$ to be well defined.

Hyper-discontinuous functions cannot be defined on the interval $[a,b]$ but we can take an "almost vertical" $\ell$ in $[a,b]$ to interesect with the hyperdiscontinuous function. Note $[a,b]\cap\mathbb{Q}\cap[0,1]$ does not have to be empty.

I feel like I had to remention $f:X\to Y$ from my post to make the answer clear. It's something I learned from a former professor.

What happens if $X$ doesn't contain any intervals with positive length? Then $f|_{[a,b]}$ isn't well defined for any $a<b$, e.g. if $X=\mathbb{Q}$.

Should I then say $\left.f\right|_{\mathbb{Q}\cap[a,b]}$. I think it still make sense.

Try and salvage whatever you can. Ignore the rest.

I'm still not sure what the line $\ell'$ is. Is it a straight line or a curve? If it's a curve, how does your choice of particular curve affect the quantity $\mathcal{D}$?

Let us continue this discussion in chat.

Intersecting random non-vertical lines with your graph seems very non-general - any kind of slightly non-linear transformation of X and Y might completely change your function. I highly doubt this approach this will give you a clean solution, even if you happen to fine tune it on your specific examples.

@Eric Could you give me an example?

As written, your quantity $\mathcal D$ will be infinite for the function $f(x)=x$, which is obviously not ideal. You need to be far more rigorous in how you define your $\ell^1$ and how it relates to your quantities.

@K.Power I don't understand. My calculations give me zero for $f(x)=x$. Are you sure you are correct?

@K.Power The most a vertical line in $[a,b]$ can interesect with $\left. f\right|_{[a,b]}$ is once. The least is also one. Therefore, $((1-1)+(1-1))/2=0$

Yes. $\mathcal D^+$ is simply the supremum over all intervals $[a,b]$ and all none-vertical lines $\ell^1$. Take the interval $[-1,1]$ and $\ell^1$ the non-vertical line defined by $y=x$. Clearly $G[-1,1]=\mathcal G[-1,1]=\{(x,x):x\in[-1,1]$. Thus the cardinality of their intersection is the continuum, and the supremum has to be larger than this.

@Arbuja You do not use a vertical line anywhere in your definition of $\mathcal D^+$

I stated the angle between the non-vertical and the vertical line approaches zero.

@Arbuja This is a non-rigorous statement. You need to define what you mean by this explicitly and where exactly this limiting procedure occurs in your definition of your various quantities. As you have posed it these quantities are not well-defined.

I don't know how to define it explcitly. All I know is when the angle between a non-vertical line and vertical line approaches zero, the most number of times a near-vertical line can intersect with a hyper-discontinuous function is infinite. The least number of times a near-veritcal line can intersect with a hyper-discontinuous functions is once.

@K.Power How would you change the defintion to get what I want?

@K.Power If you can add an answer changing my definition, I will give you the bounty.

@[email protected] I made edits.