Let E be a subset in R, f a real-value function on E.
Prove that f is continuous on E⟺ for every open subset V of R, f−1(V) in open relative to E.
My question is about the (⇒) direction only.
Let f be a continuous function on E and V a open subset on R.
If f−1(V)={}, then it is open. Suppose that f−1(V)≠{}. Let p∈f−1(V).
Then f(p)∈V. Select ϵ such that Nϵ(f(p))⊂V.
My question is this. At this point, we do not know if p is an element of E.
If p∈E, since f is continuous on E, ∃δ such that f(x)∈Nϵ(f(p)) for all x∈Nδ(p)∩E.
Thus Nδ(p)∩E⊂f−1(V).
But, suppose that p∉E. How do I know that the above statement is still true?
I tried the following:
Let q∈E be a point such that f(q)∈Nϵ(f(p))
Select α such that Nα(f(q))∈Nϵ(f(p)).
Then ∃δ such that f(x)∈Nα(f(q)) for all x∈Nδ(q)∩E.
But this only shows that Nδ(q)∩E⊂f−1(V), not Nδ(p)....
I also thought about showing that if p∉E, then Nδ(p)∩E={},
but I have no idea about how to do it.
Answer
You must change only one step in your proof:
When you say:
"If f−1(V)={} then it is open"
reeplace by
"If f−1(V)∩E={}" then f−1(V) is open relative to E".
Then the following line must be
"suppose now f−1(V)∩E≠{} then exists p∈f−1(V)∩E"
and your problem was solved since p∈E.
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