Show that $(\ell^p, d_p)$ is not a metric space for $p<1.$ Here $\ell^p$ and $d_p$ are similarly defined as $$\ell^p = \left\{ x: \mathbb{N} \to \mathbb{R} \Big| \sum_{i=1}^\infty |x_i|^p < \infty \right\}, \quad d_p(x,y) = \left( \sum_{i=1}^\infty |x_i - y_i|^p\right)^{\frac{1}{p}}. $$
Assume that $p=-1$, $$d_{-1}(x,y) = \left( \sum_{i=1}^\infty |x_i - y_i|^{-1}\right)^{-1} \quad \forall x, y \in \ell^{-1}$$ and the definiteness axiom is not met: $d_{-1}(x, y)= 0 \iff x=y$. This is because, if $d_{-1}(x,y) = 0$, then the function has a zero denominator which is undefined and $(\ell^{-1}, d_{-1})$ is not a metric space. My question is how do we prove it generally that $(\ell^{-1}, d_{-1})$ is not a metric, what I proved is just one possible case when $p=-1.$
Your proof for $p = -1$ extends to all negative $p$. $p = 0$ is also trivial. So you need $0 < p < 1$. You might check for violations of the triangle inequality, as that is the most important part of the definition of a metric. Consider $3$ points $x$, $y$ and $z$ and fill in the formulas. Try small numbers so you can see what is going on. For example, with $p = 1/2$, try $x = (4,4,0,0,0...)$, $y = (4,0,0,0,0...)$ and $z = (0,0,0...)$. $$(\sqrt{4-4} + \sqrt{4-0} + 0...)^2 = 4=d(x,y)$$ and $$(\sqrt{4-0} + 0 + 0 + ...)^2 = 4=d(y,z)$$ $$(\sqrt{4-0} + \sqrt{4-0} + 0 + ...)^2 = 16 = d(x,z)$$
We are supposed to have $d(x,y) + d(y,z)\geq d(x,z)$ but this is not the case.