How to construct a ring $ R$ such that $(Spec(R), \tau)$ is not a Sequential Space where $\tau$ is the Zariski Topology on Spec(R).
I've just learned about Zariski topology so I really don't have any idea about to solve this problem Please help!
Added: Actually I'm trying to find a function $f : X \to Y$ which is sequential continuous but not Continuous.For this problem I've already proved that such $f$ will exist only if $X$ is a Sequential Space,also I was able to construct example by taking $ X= \omega +1$ with order topology.But I was wondering if there is a ring $R$ for which $Spec(R)$ with Zariski topology is not sequential space.