Hint: is interior if and only if is interior.
Let and be open sets in and let be a one-to-one mapping from onto (so that there is an inverse mapping ). Suppose that and are both continuous. Show that for any set whose closure is contained in we have .
I am completely lost. I am assuming that the way to go about this is to show that one is a subset of the other but other than that I have no idea how to proceed.
Edit: So here is my so far very vague attempt or at least idea of how this proof might lead to: If , then so for all and . If this is the correct start to the proof, I am guessing that the continuity of the mappings comes into play but again, I'm lost.
I assume you mean interior to their respective sets of and . So here is my attempt at showing that :
Let . Then . So is not in the interior of , so .
Could you prove your hint statement and clarify it a bit more? I'm still confused and I don't think my attempt is right or at the very least it is not rigorous.
I assume you mean "...since is open..." and if that is case then what we are using the fact that for any function , continuous and open implies open as well, correct? And so for the main proof how would we proceed? I think I have a general idea but nothing I can put down into rigorous proof.
I am just not sure what you mean by " open" since is a function, not a set. But I think I have some idea for a proof:
Let such that . So and . So is not an interior point of . Since is a continuous function, we have that is not an interior point of [tex]f(S). But , so f(x)\inso .
I don't see why . That seems like just a different version of what I'm currently trying to prove. Sorry, I'm just having a hard time understanding this material.
Okay, so I know how to prove .
Let . Then or . If , then , which implies . If , then for any , . So there is a sequence of points in that converges to . Since is continuous on , there is a sequence of points in that converges to . Thus . So in either case, , so . A similar argument leads to the conclusion that .
But now how do I use this to prove ?
Edit: Alright, here is my attempt:
Lemma: If is a bijection, then .
Proof: Suppose to the contrary that . Then there exists such that . So there are at least two distinct elements in the domain of , say , such that . Then f is not injective, which is a contradiction. Q.E.D.
So now let . Then , which means there is a sequence of points in and that converge to , and since is continuous, there is a sequence of points in and that converge to , so , so . A similar argument shows , so . Q.E.D.
Your proof is correct, though in my opinion somewhat cumbersome. Here is a way it could be improved (in my opinion).
Since both and are continuous, induces two bijections : (1) a bijection between the closed sets with and the closed sets with , and (2) a bijection between the open sets contained in and the open sets contained in . One way to characterize the closure of a set is as the intersection of all closed sets containing , and one way to characterize the interior is as the union of all open sets contained in . So we have
So we have .
The definition may not refer to any other representation dependent structure besides that.
A homeomorphism preserves these predicates and therefore all properties that are definable with these predicates.
For example the boundary of a set is equivalent to this definition
X is the boundary of S
X is the largest set, such that
for all open U:
One could translate this definition into a formula of predicate logic , which uses only the above mentioned predicates. Homeomorphisms preserve such formulas, that means if f is homeomorphism and is true, then also is true.
So if X is the boundary of S, then f(X) is the boundary of f(S).
All the books I have seen define "topological property" as a "property preserved by homeomorphisms". It's clear, as you mention, that all properties definable within the predicate logic you describe are topological properties. How would you show the reverse inclusion? (My knowledge of logic is very limited!)