My teacher defined a metric space $(M,d)$ to be a set $M$ equipped with a function $d:M\times M\to \mathbb{R}$ which must satisfy a number of inequalities/properties. From here he says the diameter of a subset $\emptyset\subset Q\subseteq M$ is equal to $\sup_{x,y\in Q}d(x,y)$, then he says "$X$ is bounded if it has a finite diameter".
However we know $d(x,y)\in \mathbb{R}$ for every $x,y\in Q$ since by definition the codomain of $d$ is the reals, and every real number is finite, therefore in order for $\sup_{x,y\in Q}d(x,y)$ to even exist it must be finite, so isn't saying it has a finite diameter redundant? If one assumes the diameter can take on values that are not finite, then what does that even mean? The diameter can be a transfinite cardinal? Shouldn't that mean the metric $d$ can map to cardinals? In which case it can't have a codomain equal to $\mathbb{R}$? Ignoring that wouldn't it still mean we have to employ cardinal arithmetic in order for say the triangle equality to make sense in this context? This doesn't seem right. I feel like what he is meaning to say is that a metric space is bounded on some set if and only if its metric function is bounded on pairs of values in that set, in which case you would define the notion of bounded sets first and then define the notion of diameter for only bounded metric spaces allowing one to avoid the previous problems.
Sometimes, when $S\subset\mathbb R$ and $S\neq\emptyset$, we say that $\sup S$ is $+\infty$ when $S$ has no upper bound.
So, a (non-empty) set has finite diameter if and only if it is bounded. Otherwise, its diameter is $+\infty$.