Let \(B\) be an open subset of a topological space \(X\). Prove that a subset \(A\subset B\) is relatively open in \(B\) if and only if \(A\) is open in \(X\).
Solution
1. If \(A\) is relatively open in \(B\), then \(A\) is open in \(X\).
By definition of the subspace topology, relative openness of \(A\) in \(B\) means that there is an open set \(U\subseteq X\) such that \[ A=B\cap U. \] The set \(B\) is open in \(X\) by hypothesis, and \(U\) is open in \(X\). Therefore their intersection \(B\cap U=A\) is open in \(X\).
2. If \(A\) is open in \(X\) and \(A\subseteq B\), then \(A\) is relatively open in \(B\).
Since \(A\subseteq B\), \[ A=B\cap A. \] The second factor \(A\) is open in \(X\). Thus \(A\) has the form \(B\cap U\) for an open subset \(U=A\) of \(X\), which is exactly the definition of being open in the subspace \(B\).
3. Hence\[ A\text{ is open in }B \quad\Longleftrightarrow\quad A\text{ is open in }X. \]
Combine 1 and 2.