A proper morphism is defined as separated, of finite type and universally closed. I wonder if the requirement of being of finite type is superfluous, i.e. if being not of finite type implies not universally closed. Recall that a morphism is of finite type if and only if it is locally of finite type and quasi-compact.


In an answer to [this question](https://mathoverflow.net/questions/23337/is-a-universally-closed-morphism-of-schemes-quasi-compact) Bjorn Poonen showed that not quasi-compact implies not universally closed. 


Is it true that being not locally of finite type (plus possibly quasi-compact) implies not universally closed? 



not locally of finite type implies  not universally closed?
Let $k$ be a field, $A=k[X\_1,X\_2,\dots]$ and $I=(X\_1,X\_2,\dots)$. Then $\mathrm{Spec}(A/I^2)\to\mathrm{Spec}(k)$ is a universal homeomorphism, but not locally of finite type.


added in edit: In particular, there is no purely topological condition which implies locally finite type.


