In \(X=\{1,2,3,4\}\) with \(\mathcal{T}=\{\varnothing,\{1\},\{3\},\{1,3\},\{1,2,3\},\{1,3,4\},\{1,2,3,4\}\}\), of which points of \(X\) is \(\{1,2\}\) a neighborhood?
The open sets inside \(\{1,2\}\) are \(\varnothing\) and \(\{1\}\); only \(1\) lies in one of them. The point \(2\) would need an open set containing it inside \(\{1,2\}\), and there is none.
A neighborhood of a point \(x\) in a topological space \((X,\mathcal{T})\) is a set \(N\subseteq X\) that contains some open set \(U\in\mathcal{T}\) to which \(x\) belongs.
\[N\ \text{neighborhood of }x\iff \exists\,U\in\mathcal{T}:\ x\in U\subseteq N\]