If \(X, Y\in \mathcal{B}\) and \(T:X\to Y\) is a bijective continuous operator, then \(T^{-1}\) is continuous and thus \(T\) is a homeomorphism.
If \(X, Y\in \mathcal{B}\) and \(T:X\to Y\) is a surjective continuous operator, then \(T\) is an open map.
If \(X, Y\in \mathcal{B}\) and \(T \in L(X, Y)\) is a closed linear operator, i.e. the graph \(\Gamma(T) \subseteq X\times Y\) is closed, then \(T\) is bounded.
Let \(X, Y\in \mathcal{B}\) and \(\left\{{ T_{\alpha}}\right\} \subseteq L(X, Y)\) be a family of uniformly pointwise bounded operators, so for all points \(x\) there exists a constant \(C_x\) such that \({\left\lVert {T_{\alpha}x} \right\rVert} \leq C_x\) for all \(\alpha\). Then there exists a constant bound that is uniform in \(x\), i.e. a \(C\) such that \({\left\lVert {T_{\alpha}x} \right\rVert}\leq C\) for all \(x\).