# Is this statement by my real analysis teacher redundant/incorrect?

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.

• Commonly the supremum of an unbounded set is taken to be the symbol $\infty$, and with this interpretation you have no disagreement with your teacher. – Stephen Aug 22 '18 at 23:20
• @Stephen But the supremum of an unbounded set doesn't exist, so how can it take on any value. Also can you help me when you say $\infty$ do you mean that you are taking a extension of the strict order $(\mathbb{R},<)\to (\mathbb{R}\cup \{\infty\},<)$ and then defining the set $\infty$ so that $x<\infty$ for all $x\in \mathbb{R}$. – david62225 Aug 22 '18 at 23:21
• Consider $Q = M = \mathbb{R}$ with the absolute value metric. – none Aug 22 '18 at 23:21

## 1 Answer

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 $+\infty$.

• What is meant by $+\infty$ is that an artificially maximum element added to the strict order $(\mathbb{R},<)$ to form $(\mathbb{R}\cup\{+\infty\},<)$ where the only defining property of the set $+\infty$ is $x<\infty$ for every $x\in \mathbb{R}$. – david62225 Aug 22 '18 at 23:26
• @david62225 I don't think it is artificial. It looks quite natural to me. But you are right about the order part. – José Carlos Santos Aug 22 '18 at 23:28
• You're right thats subjective, sorry its hard when questioning a definition to not sound a little hostile. – david62225 Aug 22 '18 at 23:29