So I have f: R -> R be continuous. Let S={x in R: f(x)=0} be nonempty. Prove S is a closed set.
So far I've started with:
Let {x_n} be a sequence in S such that x_n -> x ?? yes? where to next??
Any help would be greatly appreciated!
Problem: Let be continuous. Define . Prove that is closed.
Proof: Let be a limit point of . Then there exists some sequence such that and . But then . And since was continuous as this means that . Therefore . Thus contains all its limit points, which consequently means it's closed.
Here is another way to do this problem
Problem: Let be continuous. Define . Prove that is closed.
Proof: Let be a limit point of . Then there exists some such that . Because is continous at we know that there exists some such that if
. Now assume that . Then . But this is a contradiction, for since one may let in the above. But no matter
what one picks there will be a point such that , but clearly . Therefore, and the conclusion follows.
Remark: This can be generalized greatly to much more interesting forms, but I digress. Also, these are the times where one wishes they were a topologist. For topologists define a function continuous if whenever a set is open(closed) in the codomain it's inverse image is open(closed) in the domain. And since is closed by definition is closed.