littleHilbert
Sep7-09, 08:52 AM
Hello, could you please check if the reasoning is correct. This is not a homework, just a part of an exercise in a book I'm reading at the moment.
Suppose X is a set, \mathcal{B}:=\{S\subset{}X:\bigcup{}S=X\}, \\
\mathcal{T}:=\{U\subset{}X:U=\bigcup_{\alpha\in{}A }\bigcap_{i\in{}I_\alpha}S_i\}, where I_\alpha is a finite index set, and A is an arbitrary index set.
(b) It is asked to prove that \mathcal{T} is the smallest topology for which all the sets in \mathcal{B} are open, i.e. \mathcal{T}=\bigcap_{\tau\supset\mathcal{B}}\tau is the intersection of all topologies containing \mathcal{B}.
Let U be an open set, contained in every topology, where S\in\mathcal{B} is among open sets. Then being a subset of X the set U\subset\bigcup_{S\in\mathcal{B}}S, so every x\in{}U is in some S_{\beta}\in\mathcal{B}. By assumption every finite intersection of sets in \mathcal{B} is open. Intersecting S_\beta with finitely many elements in \mathcal{B} that contain x will give us the open subset \bigcap_{S\in\mathcal{B}'\subset\mathcal{B}}S\subs eteq{}S_\beta. Taking the union \bigcup_{x\in{}U}\bigcap_{S\in\mathcal{B}'\subset\ mathcal{B}}S we thus obtain an open (in each \tau) set, which is of the form given by \mathcal{T}.
Conversely, let U\in\mathcal{T} be open. We need to show that U is open in every topology \tau, containing \mathcal{B}. For some \alpha\in{}A an element x\in{}U is contained in V:=\bigcap_{i\in{}I_\alpha}S_i. But on the other hand V is open in \tau and is a subset of U. This means that every point x\in{}U is contained in a subset of U that is open in \tau. Hence U is a union sets open in \tau and is thus open in \tau.
Suppose X is a set, \mathcal{B}:=\{S\subset{}X:\bigcup{}S=X\}, \\
\mathcal{T}:=\{U\subset{}X:U=\bigcup_{\alpha\in{}A }\bigcap_{i\in{}I_\alpha}S_i\}, where I_\alpha is a finite index set, and A is an arbitrary index set.
(b) It is asked to prove that \mathcal{T} is the smallest topology for which all the sets in \mathcal{B} are open, i.e. \mathcal{T}=\bigcap_{\tau\supset\mathcal{B}}\tau is the intersection of all topologies containing \mathcal{B}.
Let U be an open set, contained in every topology, where S\in\mathcal{B} is among open sets. Then being a subset of X the set U\subset\bigcup_{S\in\mathcal{B}}S, so every x\in{}U is in some S_{\beta}\in\mathcal{B}. By assumption every finite intersection of sets in \mathcal{B} is open. Intersecting S_\beta with finitely many elements in \mathcal{B} that contain x will give us the open subset \bigcap_{S\in\mathcal{B}'\subset\mathcal{B}}S\subs eteq{}S_\beta. Taking the union \bigcup_{x\in{}U}\bigcap_{S\in\mathcal{B}'\subset\ mathcal{B}}S we thus obtain an open (in each \tau) set, which is of the form given by \mathcal{T}.
Conversely, let U\in\mathcal{T} be open. We need to show that U is open in every topology \tau, containing \mathcal{B}. For some \alpha\in{}A an element x\in{}U is contained in V:=\bigcap_{i\in{}I_\alpha}S_i. But on the other hand V is open in \tau and is a subset of U. This means that every point x\in{}U is contained in a subset of U that is open in \tau. Hence U is a union sets open in \tau and is thus open in \tau.