There is a very useful way of thinking of continuity in terms of limits of sequences found in
the following theorem. In words, it says a function is continuous if it takes convergent
sequences to convergent sequences whenever possible.
Theorem 6.1.1A function f : D
(f)
→ Fiscontinuous at x ∈ D
(f)
if andonly if, whenever x_{n}→ x with x_{n}∈ D
(f)
, it follows f
(xn)
→ f
(x)
.
Proof: Suppose first thatf is continuous at x and let x_{n}→ x. Let ε > 0 be given.
By continuity, there exists δ > 0 such that if
|y − x |
< δ, then
|f (x)− f (y)|
< ε.
However, there exists n_{δ} such that if n ≥ n_{δ}, then
|xn − x|
< δ and so for all n this
large,
|f (x)− f (xn)| < ε
which shows f
(xn)
→ f
(x)
.
Now suppose the condition about taking convergent sequences to convergent sequences
holds at x. Suppose f fails to be continuous at x. Then there exists ε > 0 and x_{n}∈ D
(f )
such
that
|x − xn|
<
1
n
, yet
|f (x )− f (xn)| ≥ ε.
But this is clearly a contradiction because, although x_{n}→ x, f
(x )
n
fails to converge to f
(x)
.
It follows f must be continuous after all. ■
Theorem 6.1.2Suppose f : D
(f)
→ ℝ iscontinuous at x ∈ D
(f)
and supposef
(xn )
≤ l
(≥ l)
where
{xn}
is a sequence of points of D
(f)
which converges to x. Thenf
(x )
≤ l
(≥ l)
.
Proof: Since f
(xn)
≤ l and f is continuous at x, it follows from Theorem 4.4.11 and
Theorem 6.1.1,
f (x) = lnim→∞ f (xn) ≤ l.
The other case is entirely similar. ■
The following is a useful characterization of a continuous function which ties together
many of the above conditions. I am being purposely vague about the domain of the function
and its range because this theorem is a general result which holds whenever it makes sense.
Theorem 6.1.3Let f be a function defined on D
(f)
. The following areequivalent.
f is continuous on D
(f)
For every ε > 0 and x ∈ D
(f)
there exists δ > 0 such that if
|y− x|
< δ andy ∈ D
(f)
, then
|f (x)− f (y)|
< ε.
For every x ∈ D
(f)
, if x_{n}→ x where each x_{n}∈ D
(f)
, then f
(x)
=
lim_{n→∞}f
(xn)
.
Whenever U is open, f^{−1}
(U )
equals the intersection of an open set with D
(f)
.
Whenver C is closed, f^{−1}
(C)
is the intersection of a closed set with D
(f )
.
Proof:To say that f is continuous on D
(f)
is to say that it is continuous at every
x ∈ D
(f)
. Therefore, 1.) implies 2.). If 2.) holds then by definition f is continuous at every
x ∈ D
(f)
so it is continuous on D
(f)
. Thus the first two conditions are equivalent. These are
equivalent to the third condition by Theorem 6.1.1. Thus the first three conditions are
equivalent. Now suppose the fouth condition holds that “inverse images of open sets are
open”. Why is f continuous at x ∈ D
(f)
? This is obviously so because if you take
U = B
(f (x),ε)
, then f^{−1}
(B (f (x),ε))
= V ∩ D
(f)
. Since x ∈ V ∩ D
(f)
, it follows
that there is some δ > 0 such that B
(x,δ)
⊆ V since V is open. Therefore, if y ∈B
(x,δ)
∩D
(f)
, it follows that f
(y)
∈ B
(f (x),ε)
and so in other words, if
|x− y|
< δ with
y ∈ D
(f)
, then
|f (x)− f (y)|
< ε. Hence f is continuous at every x ∈ D
(f)
. Thus f
is continuous on D
(f)
. Conversely, suppose f is continuous on D
(f)
and let U
be open. Consider x ∈ f^{−1}
(U)
. By assumption f is continuous at x. Let ε > 0
be such that B
(f (x),ε)
⊆ U. Then by the definition of continuity, there exists
δ_{x}> 0 such that if
|y − x|
< δ_{x} and y ∈ D
(f )
, then f
(y)
∈ B
(f (x) ,ε)
. In other
words,
B(x,δx)∩ D (f ) ⊆ f−1( B (f (x),ε)) ⊆ f −1(U)
Hence
f− 1(U ) = ∪ −1 (B(x,δx)∩ D (f )) = (∪ −1 B (x,δx))∩ D (f)
x∈f (U) x∈f (U)
which is the intersection of an open set with D
(f)
. Thus the first four conditions are
equivalent. Now let C be closed and suppose condition 4.)
−1 −1( C )
f (C )∪f C = D (f)
Since every x ∈ D
(f)
either has f
(x)
∈ C or f
(x)
∈∕
C.
( ) ( ( ) )
f− 1(C)∩ D (f) ∪ f− 1 CC ∩ D(f) = D (f)
Now since C^{C} is open, there is an open set V such that
−1 ( C)
f C ∩ D (f) = V ∩ D (f)
Hence f^{−1}
(C )
∩ D
(f)
= V^{C}∩ D
(f)
. Thus 4.) implies 5.). Now assume 5.). Then if C
is closed, there is a closed set H such that f^{−1}
(C)
∩ D
(f)
= H ∩ D
(f)
. Then
H^{C}∩D
(f)
= f^{−1}
( )
CC
∩D
(f)
which is an open set intersected with D
(f)
. It follows, since
a generic open set is of the form C^{C}, that 4.) holds. ■