Consider $X=\{1,2,3\}$ and a topology
$$U=\{∅,\{1\},\{1,2\},\{1,2,3\}\}$$
Is $(X,U)$ a Hausdorff space?
My solution:
Points $x$ and $y$ in a topological space $X$ can be separated by neighbourhoods if there exists a neighbourhood $U$ of $x$ and a neighbourhood $V$ of $y$ such that $U$ and $V$ are disjoint ($U ∩ V = ∅)$.
$X$ is a Hausdorff space if all distinct points in $X$ are pairwise neighborhood-separable.
So by the defintion we pick element 1 and 2. The "smallest" neighbourhood of 1 is $\{1\}$, and of 2 is $\{1,2\}$. And intersection isn't the empty set. So this isn't true.
My question is this approach ok? Can we say that neighbourhood of 1 is $\{1\}$?
Best Answer
Your approach is fine. $\{1\}$ is definitely a neighborhood of $1$ (that's what it means when $\{1\}$ appears in the definition of $U$ like that), and in this case each element actually has a smallest neighborhood, so it suffices to use those to prove / disprove Hausdorff-ness.