Workshop 4
David Green
July 2025
1 (a) (b) (c):
Note that the following proof solves parts a, b, and c simultaneously.
We’ll directly prove that there exists a point x∗ ∈ F s.t. |x∗ − p| = dist(F, p), which implies
that the constant sequence {x∗ }∞
n=1 ⊆ F is bounded and converges to dist(F, p).
If x∗ is a limit point of F then x∗ ∈ F since F is closed.
All that’s left to show is that if x∗ is not a limit point of F then x∗ is also in F .
Proof that x∗ is not a limit point of F =⇒ x∗ ∈ F :
There are two possiblities, dist(F, p) = p − x∗ or dist(F, p) = x∗ − p
(Note: For the sake of brevity we’ll only show the first case but note that the same logic can be
used for the second case as well)
Case 1: dist(F, p) = p − x∗ :
∀ε > p − x∗ (Bε (p) ∩ F ) ̸= ∅.
x is not a L.P. of F =⇒ ∃ ε∗ > 0 s.t. (Bε∗ (x∗ ) ∩ F )\{x∗ } = ∅
Let FG = {x ∈ F | x > p} ⊆ F .
If FG = ∅, then let ε = p − (x∗ − ε∗ ), otherwise let ε = min(p − (x∗ − ε∗ ), dist(FG , p)).
Thus (Bε (p) ∩ F ) ̸= ∅.
Since Bε (p) = (p − ε, x∗ ) ∪ {x∗ } ∪ (x∗ , p + ε), we can say that
((p − ε, x∗ ) ∩ F ) ∪ ({x∗ } ∩ F ) ∪ ((x∗ , p + ε) ∩ F ) ̸= ∅.
What we’re gonna do is show that ((p − ε, x∗ ) ∩ F ) = ∅ and ((x∗ , p + ε) ∩ F ) = ∅,
which implies that ({x∗ } ∩ F ) ̸= ∅ =⇒ x∗ ∈ F
ε ≤ p − (x∗ − ε∗ ) =⇒ (p − ε, x∗ ) ⊆ (p − (p − (x∗ − ε∗ )), x∗ ) = (x∗ − ε∗ , x∗ ) ⊆ Bε∗ (x∗ )
Since (Bε∗ (x∗ ) ∩ F )\{x∗ } = ∅, ((p − ε, x∗ ) ∩ F ) = ∅.
We know that (x∗ , p] ∩ F = ∅
If FG = ∅ then (p, p + ε) ∩ F = ∅
If FG ̸= ∅, then we can say that since ε ≤ dist(FG , p), (p, p + ε) ∩ F = ∅.
Thus, ((x∗ , p + ε) ∩ F ) = ∅.
1
2:
=⇒ :
Suppose lim f (x) = L. Then ∀ε > 0 ∃ δ > 0 s.t. 0 < |x − x0 | < δ =⇒ |f (x) − L|.
x→x0
x − x0 > 0 =⇒ |x − x0 | = x − x0 and x − x0 < 0 =⇒ |x − x0 | = −(x − x0 ) = x0 − x.
Thus we get that ∀ε > 0 ∃ δ > 0 s.t. 0 < x − x0 < δ =⇒ |f (x) − L|
and ∀ε > 0 ∃ δ > 0 s.t. 0 < x0 − x < δ =⇒ |f (x) − L|
⇐=:
Suppose lim+ f (x) = L and lim− f (x) = L.
x→x0
x→x0
Then ∀ε > 0 ∃ δ > 0 s.t. 0 < x − x0 < δ =⇒ |f (x) − L|
and ∀ε > 0 ∃ δ > 0 s.t. 0 < x0 − x < δ =⇒ |f (x) − L|
0 < x − x0 =⇒ x − x0 = |x − x0 |
0 < x0 − x =⇒ 0 > x − x0 =⇒ |x − x0 | = −(x − x0 ) = x0 − x
Thus, ∀ε > 0 ∃ δ > 0 s.t. 0 < |x − x0 | < δ =⇒ |f (x) − L|
3:
(a)
We’ll first show that i =⇒ ii:
Suppose f : R → R is continuous.
Let U ⊆ R be open.
Let x0 ∈ f −1 (U ). Then f (x0 ) ∈ U .
Since U is open, ∃ ε∗ > 0 s.t. Bε∗ (f (x0 )) ⊆ U
Since f is continuous at any x0 ∈ R we can say
∀ε > 0 ∃ δ > 0 s.t. ∀x ∈ R (|x − x0 | < δ =⇒ |f (x) − f (x0 )| < ε =⇒ f (x) ∈ Bε (f (x0 ))). Thus
if we let ε = ε∗ we get that
∃ δ > 0 s.t. ∀x ∈ R (|x − x0 | < δ =⇒ |f (x) − f (x0 )| < ε =⇒ f (x) ∈ Bε (f (x0 )) ⊆ U =⇒
f (x) ∈ U =⇒ x ∈ f −1 (U )), giving us that
∃ δ > 0 s.t. (∀x ∈ Bδ (x0 )(x ∈ f −1 (U ) =⇒ Bδ (x0 ) ⊆ f −1 (U )) =⇒ f −1 (U ) is open.
Next we’ll show that ii =⇒ iii:
Let F ⊆ R be closed. Then F c is open.
Thus f −1 (F c ) = (f −1 (F ))c is open =⇒ f −1 (F ) is closed.
Finally, we’ll show that iii =⇒ i:
We’ll use the topological definition of continuity which states that a function f is continuous
if for every open U ⊆ R, f −1 (U ) is open.
Suppose for every closed set F ⊆ R, f −1 (F ) is closed.
Let U ⊆ R be open.
Then U c is closed =⇒ f −1 (U c ) = (f −1 (U ))c is closed =⇒ f −1 (U ) is open.
2
(b)
i.
False. f (x) = 0 is continuous (trivially) since ∀ε > 0(|f (x) − f (x0 )| = 0 < ε),
however for any U ⊆ R, f (U ) = {0}, which is not open.
ii
False. Let f (x) = tan−1 (x). f is continuous but if we let F = R which is closed, then f (F ) = ( π2 , π2 )
which is not closed.
3