Let X = \(\mathbb{R}\) with usual topology. Then the closure of (0, 1) in \(\mathbb{R}\) equals [0, 1] as [0, 1] is the smallest closed set containing (0, 1). However, the closure of (0, 1) in the subspace topology on [0, 1) equals [0, 1) as [0, 1) is the smallest closed in the subspace topology that contains (0, 1).

- Voclasses