Consider the set of natural numbers $\Bbb N$ endowed with the topology $\tau$ given by $\tau = \{\{1 \},\{2 \},\{1,2 \},\{1,2,3 \},\cdots \}$. Is $(\Bbb N, \tau)$ a compact topological space?
I have tried to proceed by taking an open cover of $\Bbb N$. Since each element in that open cover is an element of $\tau$ and every element of $\tau$ contains at most finitely many elements so there doesn't exist any finite subcollection of that cover which covers $\Bbb N$. I think this is the reason for which $(\Bbb N, \tau)$ is not compact.
Is my reasoning correct at all? Someone please verify it.
Thank you very much.