Math 313: Lecture 22
M. K. Vemuri
Limits
Definition
Let (X , dX ) and (Y , dY ) be metric spaces, E ⊆ X , f : E → Y ,
and p a limit point of E . Let q ∈ Y . We write
lim f (x) = q,
x→p
if for all ε > 0 there exists δ > 0 such that for all x ∈ E ,
0 < dX (x, p) < δ =⇒ dY (f (x), q) < ε.
Note that p need not be a point of E . Moreover, even if p ∈ E , it
may not be the case that limx→p f (x) = f (p).
Limits (contd)
Theorem
Let X , Y , E , f , p and q be as in the above definition. Then
lim f (x) = q
x→p
iff
lim f (pn ) = q
n→∞
for every sequence {pn } in E such that pn ̸= p for all n and
limn→∞ pn = p.
Proof. Assume limx→p f (x) = q, and {pn } is a sequence in E
such that pn ̸= p for all n and limn→∞ pn = p. Let ε > 0 be given.
Then there exists δ > 0 such that dY (f (x), q) < ε if x ∈ E and
0 < dX (x, p) < δ, and there exists N such that n ≥ N implies
0 < d(pn , p) < δ. Thus n ≥ N implies dY (f (pn ), q) < ε. It follows
that limn→∞ f (pn ) = q.
Limits (contd)
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Conversely, assume that it is not the case that limx→p f (x) = q.
Then there exists ε > 0 such that for every N ∈ N, there is a point
pn ∈ E , with 0 < dX (pn , p) < 1/n and d(f (pn ), q) ≥ ε. Then
pn ̸= p for all n, and pn → p, but {f (pn )} does not converge to
q.
Corollary
If f has a limit at p, this limit is unique.
This follows from the previous theorem and the fact that limits of
sequences are unique.
Continuous functions
Definition
Suppose (X , dX ) and (Y , dY ) are metric spaces, E ⊆ X , p ∈ E ,
and f : E → Y . Then f is said to be continuous at p if for all
ε > 0 there exists δ > 0 such that for all x ∈ E
dX (x, p) < δ =⇒ dY (f (x), f (p)) < ε.
If f is continuous at every point of E , then f is said to be
continuous on E .
Note that f has to be defined at p in order to be continuous at p.
If p is an isolated point of E , then the condition of continuity at p
is vacuous, and every function defined on E is continuous at p.
Continuous functions (contd)
Theorem
In the situation of the previous definition, assume also that p is a
limit point of E . Then f is continous at p iff limx→p f (x) = f (p)
The proof is a matter of carefully comparing the definitions of limit
and continuous function, and is left to the reader.
Theorem
Suppose X , Y , Z are metric spaces, E ⊆ X , p ∈ E , f : E → Y ,
g : f (E ) → Z and h = g ◦ f . If f is continuous at p and g is
continuous at f (p) then h is continous at p.
Proof. Let ε > 0 be given. Since g is continuous at f (p), there
exists η > 0 such that for all y ∈ f (E )
dY (y , f (p)) < η =⇒ dZ (g (y ), g (f (p))) < ε.
Since f is continuous at p, there exists δ > 0 such that for all
x ∈E
dX (x, p) < δ =⇒ dY (f (x), f (p)) < η.
Continuous functions (contd)
Therefore for all x ∈ E
dX (x, p) < δ =⇒ dZ (g (f (x)), g (f (p))) < ε.
It follows that h is continous at p.
Theorem
Let (X , dX ) and (Y , dY ) be metric spaces and f : X → Y . The
function f is continuous (on X ) iff f −1 (V ) is open in X for every
open set V in Y .
Proof. Assume f is continuous (on X ) and V is open in Y . Let
p ∈ f −1 (V ). Since V is open there exists ε > 0 such that
Nε (f (p)) ⊆ V . Since f is continuous at p, there exists δ > 0 such
that f (Nδ (p)) ⊆ Nε (f (p)), whence f (Nδ (p)) ⊆ V , whence
Nδ (p) ⊆ f −1 (V ). Therefore p is an interior point of f −1 (V ). It
follows that f −1 (V ) is open.
Continuous functions (contd)
Conversely, assume that f −1 (V ) is open in X for every open set V
in Y . Fix p ∈ X and let ε > 0 be given. Since Nε (f (p)) is open in
Y , it follows that U = f −1 (Nε (f (p))) is open in X . Since p ∈ U,
it is an interior point of U, and so there exists δ > 0 such that
Nδ (p) ⊆ U. Therefore f (Nδ (p)) ⊆ Nε (f (p)), i.e.
dX (x, p) < δ =⇒ dY (f (x), f (p)) < ε.
It follows that f is continous at p. Since p is arbitrary, it follows
that f is continuous (on X ).
Corollary
A mapping f of a metric space X into a metric space Y is
continuous iff f −1 (C ) is closed in X for every closed set C in Y .
https://chatgpt.com/c/67f9690e-fa38-8012-8555-2ce2b1c821f9
Continuity and Compactness
Theorem
Suppose f is a continuous mapping of a compact metric space X
into a metric space Y . Then f (X ) is compact.
Proof.
Let {Vα } be an open cover of f (X ). Since f is continuous,
{f −1 (Vα )} is an open cover of X . Since X is compact, there exist
α1 , . . . , αn such that
X ⊆ f −1 (Vα1 ) ∪ · · · ∪ f −1 (Vαn ).
Since f (f −1 (E )) ⊆ E for every E ⊆ Y , it follows that
f (X ) ⊆ Vα1 ∪ · · · ∪ Vαn .
Some homework
Read articles 4.3 – 4.4 (p85), 4.9 – 4.13 (p87-89)
