Can we extend any metric space to any larger set? Let $(X,d)$ be metric space and $X\subset Y$. Can $d$ be extended to $Y^2$ so that $(Y,d)$ is a metric space?
Edit:
how about extending any $(\Bbb Z,d)$ to $(\Bbb R,d)$

Let $Z=Y\setminus X$. Let $\kappa=|X|$. If $|Z|\ge\kappa$, we can index $Y=\{z(\xi,x):\langle\xi,x\rangle\in\kappa\times X\}$ in such a way that $z(0,x)=x$ for each $x\in X$. Now define
$$\overline d:Y\times Y\to\Bbb R:\langle z(\xi,x),z(\eta,y)\rangle\mapsto\begin{cases}
d(x,y),&\text{if }\xi=\eta\\
d(x,y)+1&\text{if }\xi\ne\eta\;.
\end{cases}$$
Then $\overline d$ is a metric on $Y$ extending $d$.
If $|Z|<\kappa$, you can still use the same basic idea. Let $\varphi:Z\to X$ be any injection, and define
$$\overline d:Y\times Y\to\Bbb R:\langle x,y\rangle\mapsto\begin{cases}
d(x,y),&\text{if }x,y\in X\\
d((\varphi(x),\varphi(y)),&\text{if }x,y\in Z\\
d(x,\varphi(y))+1,&\text{if }x\in X\text{ and }y\in Z\\
d(\varphi(x),y)+1,&\text{if }x\in Z\text{ and }y\in X\;.
\end{cases}$$
