Does there exist a set that contains all metric spaces?

Question

A metric space is formally defined as a pair (not necessarily ordered) $(X, d)$ such that $X$ is a set and $d$ is a metric.

So it got me thinking, could there exist a set that contains all metric spaces? Supposing that this was true, then, as the discrete metric is a metric for all sets, it must contain the metric space with the discrete metric... But I can't seem to understand why this would/could lead to a contradiction...

How would you answer this... Thank you!

Is there a set that contains all sets? – Math1000 Dec 23 '19 at 07:48 — Math1000, Dec 23 '19 at 07:48
No, but that argument may not work here... – PCeltide Dec 23 '19 at 08:07 — PCeltide, Dec 23 '19 at 08:07

score 2 · Accepted Answer · answered Dec 23 '19 at 10:55

Russell’s paradox works here much as in the original version; we just have to replace “contains itself” by “contains itself as a metric space”.

Let $M$ be the set of all metric spaces. Then $(M,d)$ with $d$ the discrete metric on $M$ is a metric space; hence $(M,d)\in M$. By restricted comprehension, $R=\{m\in M\mid \not\exists(X,d)\in m:X=m\}$ is a set. Now if $(M,d)\in R$, then $(M,d)\not\in R$, and if $(M,d)\not\in R$, then $(M,d)\in R$.

Does there exist a set that contains all metric spaces?

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