The document discusses continuity and uniform continuity in functions, presenting definitions and examples to illustrate these concepts. It explains continuity at a point, uniform continuity as applicable to entire sets, and homeomorphisms in topology. Theorems are also provided, establishing conditions under which continuity leads to uniform continuity on closed intervals.

1. Continuity
Homeomorphism
Uniform continuity
Unit 3

2. continuity
 A function is continuous at a point c in its
domain D if: given any > 0 there exists a > 0 such
that: if x D and | x - c | < then | f(x) - f(c) | < A
function is continuous in its domain D if it is
continuous at every point of its domain.

3. Example -1
 If f is continuous at a point c in the domain D,
and { xn } is a sequence of points in D converging to c,
then f(x) = f(c).If f(x) = f(c) for every
sequence { xn } of points in D converging to c, then f is
continuous at the point c.

4. Example -2
If f : (a, b) → R is defined on an open interval, then f
is continuous on (a, b) if and only if limx→c f(x) =
f(c) for every a < c < b since every point of (a, b) is an
accumulation point. Example 3.4. If f : [a, b] → R is
defined on a closed, bounded interval, then f is
continuous on [a, b] if and only if limx→c f(x) = f(c)
for every a < c < b, lim x→a+ f(x) = f(a), lim x→b−
f(x) = f(b2.4.

5. Example-3
Suppose that A = { 0, 1, 1 2 , 1 3 , . . . , 1 n , . . . } and
f : A → R is defined by f(0) = y0, f ( 1 n ) = yn for
some values y0, yn R. Then 1/n is an isolated point
∈
of A for every n N, so f is continuous at 1/n for
∈
every choice of yn. The remaining point 0 A is an
∈
accumulation point of A, and the condition for f to be
continuous at 0 is that limn→∞ yn = y0. As for limits,
we can give an equivalent sequential definition of
continuity, which follows immediately from Theorem

6. HOMEOMORPHISM
 . Let (X, T ) and (Y, U) be topological spaces, and let
f : X → Y be a bijection. f is said to be a
homeomorphism if f is continuous and its inverse f −1
is continuous. In this case we say that (X, T ) and (Y,
U) are homeomorphic, and write (X, T ) ' (Y, U), or
more often simply X ' Y if the topologies are
understood from context.

7. Proposition
 Let (X, T ) and (Y, U) be topological spaces, and let f :
X → Y be a bijection. Then the following are
equivalent.
 1. f is a homeomorphism.
 2. f is continuous and open
 3. f is continuous and closed. 4. U X is open if and
⊆
only if f(U) Y is open. s continuous and open
⊆

8. Example
1. Any non-constant affine linear function is a homeomorphism from
Rusual to itself. For which other topologies on R that we know is this true?
2. Let X be a set with two or more elements, and let p 6= q X. A
∈
function f : (X, Tp) → (X, Tq) is a homeomorphism if and only if it is a
bijection such that f(p) = q.
3. A function f : X → Y where X and Y are discrete spaces is a
homeomorphism if and only if it is a bijection.
4. tan : (− π 2 , π 2 ) → R is a homeomorphism. (Technically we have
never defined a topology on that open interval, but you should be able to
imagine what it looks like.)
5. There exist no homeomorphisms f : Q → Rusual, since these sets cannot
be put into bijection with one another

9. Uniform continuity
Let f : R → R be defined on S R. Then f is
⊆
uniformly continuous on S if for each > 0 there is a δ
ǫ
> 0 so that if x, y S and |x−y| < δ then |f(x) − f(y)| <
∈
. We will say that f is uniformly continuous if it is
ǫ
uniformly continuous on dom(f). Note that this says
that if f is uniformly continuous on S then for any
given > 0 the choice of δ > 0 works for the entire set
ǫ
S

10. Note that if a function is uniformly continuous on S, then
it is continuous for every point in S. By its very definition
it makes no sense to talk about a function being uniformly
continuous at a point. Now, we can show that the function
f(x) = 1/x2 is uniformly continuous on any set of the form
[a, +∞). To do this we will have to find a δ that works for a
given at every point in [a, +∞). We have f(x) − f(y) = (y
ǫ
− x)(y + x) x 2y 2 . We want to see if we can prove that the
term x + y x 2y 2 is bounded by some number M on [a,
+∞). Once we have done that we can take δ = /M. Now, x
ǫ
+ y x 2y 2 = 1 x 2y + 1 xy2 ≤ 1 a 3 + 1 a 3 = 2 a 3 . Thus,
we will take δ = a3 2 .
ǫ

11. Theorem
If f is continuous on a closed interval [a, b], then f is
uniformly continuous on [a, b].

12.  Assume that f is not uniformly continuous on [a, b].
Then there is an > 0 such that for each
ǫ δ > 0 the
implication “|x − y| < δ implies |f(x) − f(y)| < ” fails.
ǫ
Therefore, for each δ > 0 there exists at least a pair of
points x, y [a, b] such that |x − y| <
∈ δ but |f(x) − f(y)|
≥ . Thus, for each n N there exist xn, yn [a, b] so
ǫ ∈ ∈
that |xn − yn| < 1 n but |f(x) − f(y)| > .
ǫ

13.  By the Bolzano-Weierstrauss Theorem (6.14) there
exists a subsequence {xnk } {xn} that converges.
⊂
Moreover, if x0 = limk→∞ xnk , then x0 [a, b].
∈
Clearly we will also have to have that x0 = limk→∞
ynk . Since f is continuous at x0 we have f(x0) = lim
k→∞ f(xnk ) = lim k→∞ f(ynk ), so lim k→∞ [f(xnk )
− f(ynk )] = 0.

14.  Since |f(xnk )−f(ynk )| ≥ for all k, we have a
ǫ
contradiction. This leads us to conclude that f is
uniformly continuous on [a, b]. Note that in view of
this theorem the following functions are uniformly
continuous on the indicated sets: x 45 on [a, b], √ x on
[0, a], and cos(x) on [a, b].

15. THANK YOU