Suppose first that ff is continuous on a closed set SS and that AA
is a closed subset of C.C. We wish to show that f-1(A)f-1(A) is closed.
Thus, let {xn}{xn} be a sequence of points in f-1(A)f-1(A) that converges to a point c.c.
Because SS is a closed set, we know that c∈S,c∈S,
but in order to see that f-1(A)f-1(A) is closed, we need to show that c∈f-1(A).c∈f-1(A). That is, we need to show that f(c)∈A.f(c)∈A.
Now, f(xn)∈Af(xn)∈A for every n,n, and,
because ff is continuous at c,c, we have by (Reference) that f(c)=limf(xn).f(c)=limf(xn).
Hence, f(c)f(c) is a limit point of A,A, and so f(c)∈Af(c)∈A because AA is a closed set.
Therefore, c∈f-1(A),c∈f-1(A), and f-1(A)f-1(A) is closed.
Conversely, still supposing that SS is a closed set,
suppose ff is not continuous on S,S, and let cc be a point of SS at which ff
fails to be continuous. Then, there exists an ϵ>0ϵ>0 and a sequence
{xn}{xn} of elements of SS such that c=limxnc=limxn but such that
|f(c)-f(xn)|≥ϵ|f(c)-f(xn)|≥ϵ for all n.n.
(Why? See the proof of Theorem 3.4.)
Let AA be the complement of the open disk Bϵ(f(c)).Bϵ(f(c)).
Then AA is a closed subset of C.C.
We have that f(xn)∈Af(xn)∈A for all n,n,
but f(c)f(c) is not in A.A.
So, xn∈f-1(A)xn∈f-1(A) for all n,n,
but c=limxnc=limxn is not in f-1(A).f-1(A).
Hence, f-1(A)f-1(A) does not contain all of its limit points,
and so f-1(A)f-1(A) is not closed.
Hence, if ff is not continuous on S,S,
then there exists a closed set AA such that f-1(A)f-1(A) is not closed.
This completes the proof of the second half of part (1).
Next, suppose UU is an open set, and assume that
ff is continuous on U.U. Let AA be an open set in C,C,
and let cc be an element of f-1(A).f-1(A).
In order to prove thatf-1(A)f-1(A) is open, we need to show that cc belongs to the
interior of f-1(A).f-1(A).
Now, f(c)∈A,f(c)∈A,AA is open, and so there exists an
ϵ>0ϵ>0 such that the entire disk Bϵ(f(c))⊆A.Bϵ(f(c))⊆A.
Then, because ff is continuous at the point c,c, there exists a δ>0δ>0 such that
if |x-c|<δ|x-c|<δ then |f(x)-f(c)|<ϵ.|f(x)-f(c)|<ϵ.
In other words, if x∈Bδ(c),x∈Bδ(c), then f(x)∈Bϵ(f(c))⊆A.f(x)∈Bϵ(f(c))⊆A.
This means that Bδ(c)Bδ(c) is contained in f-1(A),f-1(A), and hence cc belongs to the interior of f-1(A).f-1(A).
Hence, if ff is continuous on an open set U,U, then f-1(A)f-1(A)
is open whenever AA is open.
This proves half of part (2).
Finally, still assuming that UU is open,
suppose f-1(A)f-1(A) is open whenever AA is open,
let cc be a point of S,S, and let us prove that ff is continuous at c.c.
Thus, let ϵ>0ϵ>0 be given, and let AA be the open set
A=Bϵ(f(c)).A=Bϵ(f(c)). Then, by our assumption, f-1(A)f-1(A) is an open set.
Also, cc belongs to this open set f-1(A),f-1(A), and hence cc belongs to the
interior of f-1(A).f-1(A). Therefore, there exists a δ>0δ>0 such that the
entire disk bδ(c)⊆f-1(A).bδ(c)⊆f-1(A).
But this means that if ∈S∈S satisfies |x-c|<δ,|x-c|<δ, then x∈Bδ(c)⊆f-1(A),x∈Bδ(c)⊆f-1(A),
and so f(x)∈A=Bϵ(f(c)).f(x)∈A=Bϵ(f(c)).
Therefore, if |x-c|<δ,|x-c|<δ, then
|f(x)-f(c)|<ϵ,|f(x)-f(c)|<ϵ, which proves that ff is continuous at c,c,
and the theorem is completely proved.
