graph theorems for topological spaces
Proof. Indeed, we will show, that is open. Let . Then and thus (since is Hausdorff) there exist open subsetes such that , and . Since is continuous, then is open in .
Note, that the condition implies, that . Therefore is a subset of . On the other hand this subset is open (since it is a product of two open sets) in product topology and . This shows, that every point in belongs to together with a small neighbourhood, which completes the proof.
Proposition 2. Let be a function, where are Hausdorff spaces with compact. If is a closed subset of in product topology, then is continuous.
Proof. Let be a closed set. We will show that is also closed. Consider projections
They are both continuous and thus is closed in . Since is also closed, then
Counterexample. Let denote the set of reals (with standard topology). Consider function given by and . It is obvious, that is discontinuous at , but also it can be easily checked, that is closed in . Note, that is not compact.
|Title||graph theorems for topological spaces|
|Date of creation||2013-03-22 19:15:09|
|Last modified on||2013-03-22 19:15:09|
|Last modified by||joking (16130)|