Analysis Solutions CIV

Math 441: Chapter IV, Continuous
Functions
Leonardo Di Giosia
(Problem 3)
Let (E, d), (E , d ) be metric spaces, f : E → E and A, B ⊂ E be closed subsets
such that E = A ∪ B.
Claim: Then if f is continuous in both A and B then it is continuous in E.
Proof. Let x ∈ E. Let {xn}∞
n=1 be some sequence in E which converges to x.
Case I : x ∈ A and x /∈ B. So x is an adherent point of A but not B. Then
we may lop oﬀ the ﬁrst few terms of the sequence {xn}∞
n=1 and have a resultant
sequence {xn}∞
n=m contained in A which converges to x. Because f is continuous
on A we know
lim
n→∞
f(xn) = f(x)
Case II: x /∈ A and x ∈ B. Then a similar argument as the above may be
worked to illustrate
lim
n→∞
f(xn) = f(x)
1

Chapter IV Continuous Functions
Case III: x ∈ A ∩ B. If there exist only a finite number of terms in either
{xn}∞
n=1 ∩ A or {xn}∞
n=1 ∩ B then as above, lopping off terms illustrates
lim
n→∞
f(xn) = f(x)
So assume both sets are infinite. Then we can partition the elements of {xn}∞
n=1
into the subsequences {xAn }∞
n=1 ⊂ A and {xBn }∞
n=1 ⊂ B, which both converge
to x. Let > 0. Then there exists an NA and a NB such that
n > NA → d (f(x), f(xAn )) <
and
n > NB → d (f(x), f(xBn )) <
So let n > max({NA, NB}). Then
d (f(x), f(xn)) <
So we have that
lim
n→∞
f(xn) = f(x)
In total, we have that f is continuous on E.
(Problem 4)
Let U and V be open intervals in R. Let f : U → V be nondecreasing onto.
Claim: Then f is continuous.
Proof. Let x ∈ U. Let > 0. Assume is small enough that b [f(x)] ⊂ V (OK
to assume because V is open). Because f is onto, there exist points x1, x2 ∈ U
such that f(x1) = f(x)− and f(x2) = f(x)+ . Let δ = min({x−x1, x2 −x}).
Let y ∈ Bδ(x). If y is greater than x then f(y) ∈ (f(x), f(x) + ) and if y is less
than x then f(y) ∈ (f(x)− , f(x)). So f(B (x)) ⊂ B (f(x)). So f is continuous
on U.
(Problem 5)
Let (E, d), (E , d ) be metric spaces. Let f : E → E , p ∈ E. Define the
oscillation of f at p to be inf({a ∈ R|∃r > 0|x, y ∈ Br(p) → d (f(x), f(y)) <
a}) = inf(S).
Claim: Then f is continuous at p if and only if the oscillation of f at p is 0.
Proof. =⇒ Assume f is continuous at p. Let > 0. Then there exists some
δ > 0 such that if x ∈ Bδ(p) we have
f(x) ∈ B2
(f(p))
2

So if x, y ∈ Bδ(p). Then by the triangle inequality
d(x, y) <
So ∈ S. So (0, ∞) ⊂ S. Assume there exists some b < 0 such that b ∈ S.
Then d (f(p), f(p)) = 0 < b < 0. So either S = (0, ∞) or S = [0, ∞), in which
case, the infemum of both sets is 0 implying that the oscillation of f at p is 0.
⇐= Assume the oscillation of f at p is 0. Then ∈ S. So there exists some
r > 0 such that if x ∈ Br(p) then d (f(s), f(p) < . So f is continuos at p if
and only if the oscillation of f at p is zero.
(Problem 6)
Let (E, d) be a metric space. Let S ⊂ E. Define f : E → R by f(x) = 1 if
x ∈ S and f(x) = 0 if x /∈ S.
Claim: Then the set of points in E at which f is discontinuous is equal to
boundary(S).
Proof. ⊂ Let x be a point of discontinuity of f. Without loss of generality
assume that x ∈ S. So x ∈ closure(S). So, we must have some sequence
{xn}∞
n=1 in E which converges to x such that
lim
n→∞
f(xn) = f(x)
So this limit must be 0, implying that there are a finite number of terms of
{xn}∞
n=1 in S. This illustrates that there are an infinite number of terms of
{xn}∞
n=1 in Sc
. So x is an adherent point of Sc
implying that x ∈ closure(Sc
).
Hence, x ∈ closure(S) ∩ closure(Sc
) = boundary(S).
⊃ Let y ∈ boundary(S). Again, assume without loss of generality that
x ∈ S. Assume that f is continuous at y. There exists some δ > 0 such that if
y0 ∈ Bδ(y) then we have
|f(y) − f(y0)| <
1
2
(Implication I)
However, because y ∈ boundary(S) we have that y ∈ closure(Sc
) implying that
y is an adherent point of Sc
so we have some element y1 ∈ Bδ(y) ∩ Sc
. So
f(y1) = 0 and f(y) = 1 directly contradicting implication I so we have that f is
discontinuous at y. In total, the set of points of discontinuity of f is equivalent
to the boundary of S.
(Problem 7)
Let (E, d) be a metric space. Let (b, c) ⊂ R be an open interval and let a ∈ (b, c).
Let f : (b, a) ∪ (a, c)c → E be a function.
Claim: Then
lim
x→a
f(x)
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exists if and only if the limits
lim
x→a+
f(x), lim
x→a−
f(x)
exist and are equal
Proof. =⇒ Let limx→a f(x) exist. Denote it q. Let {xn}∞
n=1 be any sequence
in (b, a) which converges to a. Then because {xn}∞
n=1 is also a subset of (b, c)
we must have that
lim
n→∞
f(xn) = q
so
lim
x→a+
f(x)
exists. An identical argument demonstrates that the right limit exists too and
is also equal to q.
⇐=
Let both the right and left limits exist and be equal. Denote this limit q. Let
> 0. Because both limits exist, there exist δ+
and δ−
such that
x ∈ (b, a) ∧ a − x < δ+
→ d(f(x), q) < (Implication I)
and
x ∈ (a, c) ∧ x − a < δ−
→ d(f(x), q) < (Implication II)
So let x ∈ (b, a) ∪ (a, c) such that |x − a| < min({δ+
, δ−
}). Then, regardless of
the side of a which x is on, we must have that
d(f(x), q) <
So limx→a f(x) exists if and only if limx→a+ f(x) and limx→a− f(x) exist and
are equal.
(Problem 8)
Let a, > 0 and deﬁne U = (a, ∞). Let f : U → R.
Claim:
lim
x→+∞
f(x)
exists if and only if there exists if and only if there exists an N > 0 such that
x, y > N → |f(x) − f(y)| <
Proof. =⇒ Let
lim
x→+∞
f(x) = q
exist. There must exist some δ such that for all z ∈ (0, 1
a ) such that if z < δ,
f
1
z
− q <
2
(Implication I)
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So for any two elements x, y > max({1
δ , a}) we have that 1
x , 1
y are of the above
conditions. the Triangle Inequality and Implication I assure that
|f(x) − f(y)| <
⇐=
Assume there exists some N > 0 such that if x, y > N we have
|f(x) − f(y)| <
Let x0, y0 be some elements of (0, 1
a ) such that x0, y0 < 1
N . Then we have that
1
x0
, 1
y0
∈ (a, +∞) such that 1
x0
, 1
y0
> N. So we must have that
f
1
x0
− f
1
y0
<
This implies that
lim
y→0
f
1
y
= lim
x→+∞
f(x)
exists (Which could explicitly be shown by considering that the sequence
{f( 1
n )}∞
n=1 is Cauchy in R). In total, we see that
lim
x→+∞
f(x)
exists if and only if there exists an N > 0 such that
x, y > N → |f(x) − f(y)| <
(Problem 9)
(b)
Claim:
lim
x→1
x − 1
√
x − 1
= 2
Proof. Note that for any x ∈ R, x = 1,
x − 1
√
x − 1
=
x −
√
x +
√
x − 1
√
x − 1
=
√
x + 1
Because this function is continuous at 1 we have that
lim
x→1
x − 1
√
x − 1
= lim
x→1
√
x + 1 = 2
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(Problem 13)
Let (E, d) be some compact metric space and let f : E → R be continuous.
Claim: Then f is bounded and attains a maximum value on E.
Proof. Assume f is not bounded. Then we have some sequence {xn}∞
n=1 ⊂ E
such that for all n ∈ N,
|f(xn)| > n
But because E is compact, there exists some subsequence, {xmn }∞
n=1 of
{xn}∞
n=1 which converges to some x ∈ E. So we must have that the sequence
{f(xmn
)}∞
n=1 converges to f(x). So this sequence is bounded while {f(xn)}∞
n=1
is not. This contradiction illustrates that E must be bounded.
So the set f(E) ⊂ R is bounded which implies s = sup(f(E)) exists. Because s
is an adherent point of f(E) there must exist some sequence {sn}∞
n=1 ⊂ f(E)
which converges to s. Thus there exists some {pn}∞
n=1 ⊂ E such that for all
n ∈ N,
f(pn) = sn
But because {pn}∞
n=1 ⊂ E it must have some convergent subsequence denoted
{pmn
}∞
n=1 with limit p ∈ E. So because f is continuous, the image of {pmn
}∞
n=1
converges to f(p). But this sequence is a subsequence of the already convergent
{sn}∞
n=1, implying that the limits are equal (f(p) = s = sup(F(E))). That is,
we have proved the existence of the point p ∈ E such that
f(p) = sup(f(E))
Which implies max(F(E)) exists.
(Problem 16)
Let (E, d) and (E , d ) be metric spaces. Let E be compact. Let f : E → E be
a continuous bijection.
Claim: g = f−1
: E → E is also continuous.
Proof. We will show that any closed set in E has an closed preimage under
the function g. Let C be some closed set in E. Let {yn}∞
n=1 be any convergent
sequence in f(C). Because f is a bijection, there is a unique sequence {xn}∞
n=1 in
C such that for all n ∈ N, f(xn) = yn. But because this sequence is in a compact
metric space, it must have a convergent subsequence, denoted {xmn }∞
n=1 with
limit x. Because C is closed, x ∈ C. So f(x) ∈ f(C). Because f is continuous,
the image of {xmn
}∞
n=1, {ymn
}∞
n=1, must converge to the image of x, f(x). But,
because we already declared {yn}∞
n=1 is convergent. So it must converge to the
limit of the subsequence. So it converges to f(x), a point of f(C).
So, f(C) is a closed set and f is a function that maps closed sets to closed sets.
Finally, because f is a bijection, for any set S in E, the image of S under f is
equivalent to the preimage of S under the inverse function, g. So if S is some
closed set in E, we know f(S) = g−1
(S) is closed. So g is a continuous function
from E to E.
6

(Problem 17)
Claim: The function f(x) = |x| is uniformly continuous but the function
g(x) = x2
is not uniformly continuous.
Proof. Let > 0. Let x, y ∈ R such that |x − y| < 2
. After factoring we see
that
|f(x) − f(y)| · |f(x) + f(y)| = ||x| − |y|| ≤ |x − y|
So assume that |f(x) − f(y)| ≥ . Because f is never negative on R, we know
that
|f(x) − f(y)| ≤ |f(x) + f(y)|
In total, we ﬁnd that
2
≤ |f(x) − f(y)|2
≤ |f(x) − f(y)| · |f(x) + f(y)| ≤ |x − y| < 2
So our assumption was false and we must have that
|f(x) − f(y)| <
So f is uniformly continuous. We now demonstrate that g is not uniformly
continuous. Say it is. Then because 1 > 0 there exists a η > 0 such that for
any x, y ∈ R, |x − y| < η → |g(x) − g(y)| < 1. Because η > 0, the number 3
2η is
deﬁned. Denote x0 = 3
2η + η
3 and y0 = 3
2η − η
3 . Because |x0 − y0| < η we know
that
|x0 − y0| · |x0 + y0| = |g(x0) − g(y0)| < 1
2η
3
·
3
η
< 1
2 < 1
So g is not uniformly continuous.
(Problem 18)
Let (E, d) be a metric space and denote f : E → E as the identity function on
E.
Claim: Then f is uniformly continuous.
Proof. Let > 0. For any x, y ∈ E such that d(x, y) < we know d(f(x), f(y)) =
d(x, y) < . So f is uniformly continuous where our funciton δ( ) = .
7

(Problem 19)
Let (E, d) be a metric space and let p0 ∈ E.
Claim: Then the function f : E → R deﬁned by f(p) = d(p, p0) is uniformly
continuous.
Proof. Let > 0. Let δ( ) = . Then let p and q be two points of E such that
d(p, q) < . By the reverse triangle inequality,
|d(p, p0) − d(p0, q)| ≤ d(p, q) <
|f(p) − f(q)| <
So f is uniformly continuous.
(Problem 20)
Let (E, d), (E , d ) and (E , d ) be metric spaces and let f : E → E and
g : E → E be uniformly continuous functions.
Claim: Then the composition function g ◦ f : E → E is also uniformly contin-
uous.
Proof. Let > 0. Then because g is uniformly continuous there exists some
δ > 0 such that if x, y are any points of E such that d (x, y) < δ, then
d (g(x), g(y)) <
Because δ > 0 and because f is uniformly continuous, there exists some δ such
that if a, b are elements of E such that d(a, b) < δ we must have that
d (f(a), f(b)) < δ
Let p, q be elements of E such that d(p, q) < δ . Then we know d (f(p), f(q)) <
δ. Then we know
d (g(f(p)), g(f(q))) <
d (g ◦ f(p), g ◦ f(q)) <
So g ◦ f is uniformly continuous.
(Problem 22)
Let V and V be vector spaces and let f : V → V be a linear transformation.
(a) Let f be continuous at some v ∈ V .
Claim: Then f is uniformly continuous.
8

Proof. Let > 0. Because f is continuous at v, there exists some δ > 0 such
that f(Bδ(v)) ⊂ B (f(v)). Let x and y be two vectors in V such that
x − y < δ
Then
v − (v − x + y) < δ
So, this second vector in parentheses is an element of Bδ(v). Additionally,
because f is a linear transformation, we must have the following:
f(v) − f(v − x + y) <
f(v) − f(v) + f(x) − f(y) <
f(x) − f(y) <
So f is uniformly continuous (and thus, continuous) on V .
(b) Denote S = { f(x) / ¯x |x ∈ V ∩ {0}c
}
Claim: f is continuous if and only if the set S is bounded.
Proof. =⇒
Let f be continuous. Assume the set S is not bounded. Let n be some natural
number. Then there exists some nonzero vector in V , denoted vn, such that
f(vn) / vn > n. We have now constructed a sequence of nonzero vectors in
V , {vn}∞
n=1.
Consider new sequence {wn}∞
n=1 defined by wn = 1
f(vn) · vn (remember that
f(vn) = 0, so this term is well defined). Then we have that for all natural
numbers n, f(wn) = 1. We next demonstrate that {wn}∞
n=1 converges to the
zero vector.
Let > 0. Let n be some natural number greater than 1
. Then
f(vn)
vn
> n >
1
vn
f(vn)
<
wn = wn − 0 <
So {wn}∞
n=1 converges to the zero vector. Because f is continuous, the image of
{wn}∞
n=1 should converge to the image of the zero vector, which would be the
zero vector in V . So the sequence { f(wn) }∞
n=1 must converge to 0. But for
all n, f(wn) = 1. So the sequence converges to both 1 and 0. So 0 = 1, which
provides our contradiction, and we find that S must be bounded.
⇐= Let S be bounded and let M be some positive upper bound of S. Let
{xn}∞
n=1 be some sequence of vectors which converge to the zero vector of V .
We want to show {f(xn)}∞
n=1 converges to the zero vector of V . Let > 0.
Then there exists some N such that n > N implies
xn <
M
9

But we know for all n,
f(xn) < M · xn
Let n > N. Then xn < M so M · xn < so
f(xn) <
Thus, {f(xn)}∞
n=1 converges to the zero vector of V . So limn→∞ f(xn) =
f(limn→∞ xn). So f is continuous at x = 0. Because f is continuous at one
point, it is continuous everywhere.
So f is continuous everywhere if and only if S is bounded.
(c) Let V be ﬁnite dimensional
Claim: Then f is continuous.
Proof. There exists basis of V , denoted B = {bi|1 ≤ i ≤ m}, for some m. Let
{xn}∞
n=1 be some sequence of vectors in V which converges to the zero vector
of V . Then for each natural number n we have real numbers αi
ni ∈ {1, 2 . . . m}
such that
xn = α1
nb1 + α2
nb2 + · · · + αm
n bm
So, because {xn}∞
n=1 converges to the zero vector, we have that
lim
n→∞
xn = 0
lim
n→∞
(α1
nb1 + α2
nb2 + · · · + αm
n bm) = 0
lim
n→∞
(α1
n)b1 + lim
n→∞
(α2
n)b2 + · · · + lim
n→∞
(αm
n )bm = 0
Because these basis vectors are linearly independent we must have that for each
i ∈ {1, 2, . . . , m}, limn→∞ αi
n = 0. Our next step is to illustrate that the image
of {xn}∞
n=1 under f converges to the zero vector of V . Let > 0. For each
i ∈ {1, 2, . . . , m}, there exists some Ni such that n > Ni implies
|αi
n| <
m · f(bi)
(Note that none of the basis vectors are the zero vector of V so for all i, f(bi) =
0 and the above number is deﬁned). So let N > max{Ni|1 ≤ i ≤ m}. Then
n > N implies
|α1
n| · f(b1) + · · · + |αm
n | · f(bm) <
α1
n · f(b1) + · · · + αm
n · f(bm) <
f(α1
n · b1 + · · · + αm
n · bm) <
f(xn) <
So {f(xn)}∞
n=1 converges to the zero function of V and because f(0) = 0 , we
have that f is continuous at x = 0 so by a previously established result, f is
continuous everywhere.
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(Problem 24)
Let f : [α, β] ⊂ R → R be continuous such that f(α) < f(β) and let γ be some
number such that f(α) < γ < f(β).
Claim: Then there exists some c in (α, β) such that f(c) = γ (the Interme-
diate Value Theorem).
Proof. Denote set S = {x ∈ [α, β]|f(x) ≤ γ}. Because α ∈ S we know S is
nonempty. Additionally, S is bounded above by β and by the completeness
axiom, s = sup(S) exists. So, additionally we have the sequence {sn}∞
n=1 ⊂ S
such that
lim
n→∞
sn = s
Because f is continuous we know that
f(s) = lim
n→∞
f(sn)
Because for all natural number n, sn ∈ S so f(sn) ≤ γ. Thus,
f(s) ≤ γ
Assume f(s) < γ. Then we know β = s, (this is because, if otherwise, f(s) =
f(β) > γ > f(s)). Because γ − f(s) > 0 and because f is continuous there
exists a δ > 0 such that for all x ∈ [α, β],
x ∈ bδ → |f(x) − f(s)| < γ − f(s) → f(x) < γ (Implication I)
Because s = β we may ﬁnd some x0 ∈ (s, β] such that x0 ∈ bδ(s). So (by
Implication I)
f(x0) < γ
So x0 ∈ S. But x > sup(S) = s. So we have that our assumption was false and
f(s) = γ
(Problem 26)
Let f be some real valued, one to one function, continuous on the interval [α, β]
such that f(α) ≤ f(β).
Claim: Then f([α, β]) = [f(α), f(β)].
Proof. By the intermediate vale theorem, we must have that [f(α), f(β)] ⊂
f([α, β]). So to show that [f(α), f(β)] ⊃ f([α, β]), let y ∈ f([α, β]). Then we
have some x ∈ [α, β] such that f(x) = y.
Assume y > f(β). Then x = α, β, so we may consider the intervals [α, x] and
[x, β]. See that y+f(β)
2 is a number between f(α) and y as well as a number
11

between f(β) and y. So, by the intermediate value theorem, we have that there
are points x1 and x2 in [α, x] and [x, β], respectively, such that
f(x1) = f(x2) =
y + f(β)
2
However, we know that f is one to one on the interval [α, β], so x1 = x2, which
is absurd (they came from intervals who intersect only at x, and these points
are certainly not equal to x). Thus, our assumption was false and we have that
y ≤ f(β).
If we assume that y < f(α), we may construct a similar argument with flipped
inequality signs, and using the value y+f(α)
2 to show that f is not one to one.
This illustrates again that y ≥ f(α).
So y ∈ [f(α), f(β)], and we must have that [f(α), f(β)] ⊃ f([α, β]) so
[f(α), f(β)] = f([α, β])
(Problem 27)
Let p(x) = αnxn
+· · ·+α0 be some polynomial of odd degree, where we require
αn = 0.
Claim: Then p : R → R is a function such that p(R) = R.
Proof. It will suffice to show that p(x) is not bounded from above or below. So
first, assume there exists an upper bound of p(x), denoted M.
Case I: αn > 0. Let N be some real number greater than −αj for any 1 ≤ j ≤
n−1, and greater than the quantity M −α0. Let x0 be some real number greater
than n·N
αn
. (Additionally, let x0 > 1 so we may assume that xn
0 > xn−1
0 · · · >
x2
0 > x0 > 1) Then
N · n < αnx0
N(nxn−1
0 ) < αnxn
0
N(xn−1
0 + xn−2
0 + · · · + x0 + 1) < αnxn
0
−αn−1xn−1
0 − αn−2xn−2
0 + · · · − α1x1
0 − α0 + M < αnxn
0
M < αnxn
0 + αn−1xn−1
0 + · · · + α0
M < p(x0)
So, there exists some point, x0, of R such that p(x0) > M which violates our
definition of M. So p(x) is not bounded above.
Case II: αn < 0. Then we construct a new polynomial, denoted q(x) = p(−x).
We must have that q is of odd degree with a positive nth
coefficient. So, by the
argument featured the first case, we must have that q(x) is not bounded above
by M (for any M ∈ R), so there exists some y0 such that q(y0) = p(−y0) > M.
Thus, we have that p(x) cannot be bounded above.
So, in both cases, p(x) cannot be bounded from above. We must also have that
p(x) cannot be bounded from below as well. (Explanation: assume there is one
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polynomial, b(x), of odd degree, which is bounded from below. Multiplying it
by −1 provides a polynomial of odd degree bounded from above which we’ve
shown is impossible.)
Thus, in total, p(x) is neither bounded above or below. If γ is any real number,
we must be able to ﬁnd α and β such that p(α) < γ < p(β) and by the
intermediate value theorem, because polynomials are continuous, there exists
some z0 in either (α, β) or (β, α) (whichever makes sense), such that
p(z0) = γ
So p(R) = R
(Problem 28)
Let [α, β]n
be some open interval in En
.
Claim: Then [α, β]n
is connected.
Proof. We proceed by the principle of mathematical induction. In the base case,
n = 1 and we have an open interval in R which is connected.
So for some k > 1, [α, β]k
is connected. Let U and V be any nonempty open
sets such that U ∪ V = [α, β]k+1
. We must have sets
Uk = {¯x ∈ (α, β)k
|∃y ∈ (α, β)|(¯x, y) ∈ U ⊂ (α, β)k+1
}
Vk = {¯x ∈ (α, β)k
|∃y ∈ (α, β)|(¯y) ∈ V ⊂ (α, β)k+1
}
U1 = {x ∈ (α, β)|∃¯y ∈ (α, β)k
|(¯y, x) ∈ U ⊂ (α, β)k+1
}
V1 = {x ∈ (α, β)|∃¯y ∈ (α, β)k
|(¯y, x) ∈ V ⊂ (α, β)k+1
}
We know that
Uk × U1 = U
and
Vk × V1 = V
From an exercise in Chapter III we know that these four sets are open in their
respective metric spaces. We want to show that [α, β] ⊂ U1 ∪ V1 and [α, β] ⊂
Uk ∪ Vk. So let x ∈ [α, β]. Then
(
α + β
2
,
α + β
2
, · · · ,
α + β
2
), x ∈ [α, β]k+1
So denote this element of [α, β]k+1
as ¯y. By hypothesis we know that either
¯y ∈ U or ¯y ∈ V . In which case, either x ∈ U1 or x ∈ V1 implying x ∈ U1 ∪ V1.
So
[α, β] ⊂ U1 ∪ V1
And because [α, β] is connected, we see that these sets must not be disjoint and
we have that there exists some a ∈ [α, β] such that
a ∈ U1 ∩ V1
13

Next, we want to show that [α, β]k
⊂ Uk ∪ Vk and [α, β]k
⊂ Uk ∪ Vk. So let
¯x ∈ [α, β]k
. Then
(¯x,
α + β
2
) ∈ [α, β]k+1
So denote this element of [α, β]k+1
as ¯z. By hypothesis we know that either
¯z ∈ U or ¯z ∈ V . In which case, either ¯x ∈ Uk or x ∈ Vk implying ¯x ∈ U1 ∪ V1.
So
[α, β]k
⊂ Uk ∪ Vk
And because [α, β]k
is connected, (induction hypothesis) we see that these sets
must not be disjoint and we have that there exists some ¯a ∈ [α, β]k
such that
a ∈ Uk ∩ Vk
So, in total we have that
(¯a, a) ∈ U ∩ V
So U and V cannot be disjoint and we must have that [α, β]k+1
is connected,
completing the induction step. So, by the principle of mathematical induction
we have that the closed interval
[α, β]n
is connected for any n ∈ N.
(Problem 29)
Let (E, d) be some arcwise connected metric space.
Claim: Then E is connected.
Proof. Assume E is not connected. then we have nonempty open sets U and
V such that U ∩ V = ∅ while U ∪ V = E. Because these sets are nonempty,
there exist some u ∈ U and v ∈ V . Then we have some continuous function
f : [0, 1] → E such that f(0) = u and f(1) = v.
Consider the sets f−1
(U) and f−1
(V ) in [0, 1]. They must be open sets because
f is continuous. We know 0 ∈ f−1
(U) and 1 ∈ f−1
(V ), so they’re not empty.
Let x ∈ [0, 1]. Either f(x) ∈ U or f(x) ∈ V (but not both) because U and V
are disjoint and union to form E. This implies that x cannot be in both f−1
(U)
and f−1
(V ) so these sets are disjoint. Furthermore, we know that because
f(x) must be in either U or V , that x ∈ f−1
(U) or x ∈ f−1
(V ) implying
[0, 1] ⊂ f−1
(U) ∪ f−1
(V ). So
[0, 1] = f−1
(U) ∪ f−1
(V )
Hence, f−1
(U) and f−1
(V ) are two nonempty disjoint open sets in [0, 1] which
union to form [0, 1]. So the closed interval [0, 1] is not connected. This contra-
diction illustrates that E is connected.
14

(Problem 30)
Let I ⊂ R2
where I is some closed interval. Let f : I → R be continuous.
Claim: Then f cannot be one to one.
Proof. Let x, p, q ∈ I such that f(x) < f(p) < f(q). (If there are no such three
points, then f is already not one to one. So we deal with this case). Because
any interval of R2
is connected, and by the result of problem 29, b, we must
have two continuous functions from the interval [0, 1] to I, denoted Txp and Txq
such that
Txp(0) = x
Txp(1) = p
Txq(0) = x
Txq(1) = q
Additionally, enforce that Txp([0, 1])∩Txq([0, 1]) = {x}. (AKA, the transforma-
tion of these line segments only intersect at the base point, x.)
Note that both f(x) < f(x)+f(p)
2 < f(p), f(q). By the generalized intermediate
value theorem (generalizing to real valued functions on any connected metric
space, not just closed intervals in R), because both Txp([0, 1]) and Txq([0, 1]) are
closed subspaces of I, we must have points xa ∈ Txp([0, 1]) and xb ∈ Txq([0, 1])
such that
f(xa) = f(xb) =
f(x) + f(p)
2
Because these points xa and xb are not equal to x, we must have that they are
distinct. Thus, f cannot be one to one.
15

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