supremum over closure
The theorem is clearly true for . Thus, it will be assumed that .
Since , we have .
Suppose first that . Let . Then there exists with . Since is continuous, there exists such that, for any with , we have . Since , there exists with . (Recall that if and only if every neighborhood of intersects .) Thus, . Therefore, . Hence, .
Now suppose that for some . Let . Then there exists with . Since is continuous, there exists such that, for any with , we have . Since , there exists with . Thus, . Therefore, . Hence, .
In either case, it follows that . ∎
Note that this theorem also holds for continuous functions , where is an arbitrary topological space. To prove this fact, one would need to slightly adjust the proof supplied here.
|Title||supremum over closure|
|Date of creation||2013-03-22 17:08:22|
|Last modified on||2013-03-22 17:08:22|
|Last modified by||Wkbj79 (1863)|