\(X\) is a top space and \(A\subset X\).
Relative topology: we can induce on \(A\) the topology of \(X\) when we declare to be open all the subsets of \(A\) of the form \(A\cap U\) where \(U\) is an open subset of \(X\). Similarly, closed sets are precisely those sets of the form \(A\cap Y\) where \(Y\) is closed sets in \(X\). Let \(B\subset A\) be given. Assume \(B\subset A\) is open ( resp. closed )in \(X\). It is clear that whatever \(A\) is, \(B\) is open ( resp. closed) in \(A\). Now, let \(Y\subset A\) again. Does the openness ( resp. the closedness) of \(Y\) in \(A\) imply its openness(resp. its closedness) in \(X\)? Answer: yes if \(A\) istself is open(resp. closed). In general, we may have \(Y\) open in \(A\) without being open in \(X\): take \(X = \mathbb{R}\), \( A = \left[ 0, \infty\right) \) and \(Y = \left[0, 1\right) \). Then \(Y = \left(-\epsilon, 1 \right)\cap A \), \(\epsilon>0\), is open in \(A\) but this set is clearly not open in \(X\).
More generally, for any \(B\subset A \subset X\), we have \( \overline{B}^{A} = \overline{B}^X \cap A \). If \(\Gamma\) is a base( resp. base) for \(X\), then \(\widetilde{\Gamma}:= \{ A\cap B: B\in \Gamma \}\) is a base ( resp. prebase) for \(A\)
. The relative topology is transitive: if \(B\) is a subset of a subspace \(A\), then the topology induced on \(B\) by the inclusion \( B \subset A\) is the same as the topology induced on \(B\) by the inclusion \( B \subset X \).Examples topological spaces that are subsets of \(\mathbb{R}^n\) and \( \ell_2 \): \(D^n, S^{n-1}, I^n \); the latter is the unit cube of \(\mathbb{R}^n\) defined as \(I^n := \{ x = (x_1, \dots, x_n): 0\leqslant x_i \leqslant 1\,, i = 1, \dots, n \}\)