Direct Proof
edit
Let
U
{\displaystyle {\mathcal {U}}}
be an open cover of
X
{\displaystyle X}
. Since
X
{\displaystyle X}
is compact we can extract a finite subcover
{
A
1
,
…
,
A
n
}
⊆
U
{\displaystyle \{A_{1},\dots ,A_{n}\}\subseteq {\mathcal {U}}}
.
If any one of the
A
i
{\displaystyle A_{i}}
's equals
X
{\displaystyle X}
then any
δ
>
0
{\displaystyle \delta >0}
will serve as a Delta number.
Otherwise for each
i
∈
{
1
,
…
,
n
}
{\displaystyle i\in \{1,\dots ,n\}}
, let
C
i
:=
X
∖
A
i
{\displaystyle C_{i}:=X\smallsetminus A_{i}}
, note that
C
i
{\displaystyle C_{i}}
is not empty, and define a function
f
:
X
→
R
{\displaystyle f:X\rightarrow \mathbb {R} }
by
f
(
x
)
:=
1
n
∑
i
=
1
n
d
(
x
,
C
i
)
.
{\displaystyle f(x):={\frac {1}{n}}\sum _{i=1}^{n}d(x,C_{i}).}
Since
f
{\displaystyle f}
is continuous on a compact set, it attains a minimum
δ
{\displaystyle \delta }
.
The key observation is that, since every
x
{\displaystyle x}
is contained in some
A
i
{\displaystyle A_{i}}
, the extreme value theorem shows
δ
>
0
{\displaystyle \delta >0}
. Now we can verify that this
δ
{\displaystyle \delta }
is the desired Delta number.
If
Y
{\displaystyle Y}
is a subset of
X
{\displaystyle X}
of diameter less than
δ
{\displaystyle \delta }
, then there exists
x
0
∈
X
{\displaystyle x_{0}\in X}
such that
Y
⊆
B
δ
(
x
0
)
{\displaystyle Y\subseteq B_{\delta }(x_{0})}
, where
B
δ
(
x
0
)
{\displaystyle B_{\delta }(x_{0})}
denotes the ball of radius
δ
{\displaystyle \delta }
centered at
x
0
{\displaystyle x_{0}}
(namely, one can choose
x
0
{\displaystyle x_{0}}
as any point in
Y
{\displaystyle Y}
). Since
f
(
x
0
)
≥
δ
{\displaystyle f(x_{0})\geq \delta }
there must exist at least one
i
{\displaystyle i}
such that
d
(
x
0
,
C
i
)
≥
δ
{\displaystyle d(x_{0},C_{i})\geq \delta }
. But this means that
B
δ
(
x
0
)
⊆
A
i
{\displaystyle B_{\delta }(x_{0})\subseteq A_{i}}
and so, in particular,
Y
⊆
A
i
{\displaystyle Y\subseteq A_{i}}
.
Proof by Contradiction
edit
Assume
X
{\displaystyle X}
is sequentially compact ,
A
=
{
U
α
|
α
∈
J
}
{\displaystyle {\mathcal {A}}=\{U_{\alpha }|\alpha \in J\}}
is an open covering of
X
{\displaystyle X}
and the Lebesgue number
δ
{\displaystyle \delta }
does not exist. So,
∀
δ
>
0
{\displaystyle \forall \delta >0}
,
∃
A
⊂
X
{\displaystyle \exists A\subset X}
with
d
i
a
m
(
A
)
<
δ
{\displaystyle diam(A)<\delta }
such that
¬
∃
β
∈
J
{\displaystyle \neg \exists \beta \in J}
where
A
⊂
U
β
{\displaystyle A\subset U_{\beta }}
.
This allows us to make the following construction:
δ
1
=
1
{\displaystyle \delta _{1}=1}
,
∃
A
1
⊂
X
{\displaystyle \exists A_{1}\subset X}
where
(
d
i
a
m
(
A
1
)
<
δ
1
)
{\displaystyle (diam(A_{1})<\delta _{1})}
and
¬
∃
β
(
A
1
⊂
U
β
)
{\displaystyle \neg \exists \beta (A_{1}\subset U_{\beta })}
δ
2
=
1
2
{\displaystyle \delta _{2}={\frac {1}{2}}}
,
∃
A
2
⊂
X
{\displaystyle \exists A_{2}\subset X}
where
(
d
i
a
m
(
A
2
)
<
δ
2
)
{\displaystyle (diam(A_{2})<\delta _{2})}
and
¬
∃
β
(
A
2
⊂
U
β
)
{\displaystyle \neg \exists \beta (A_{2}\subset U_{\beta })}
⋮
δ
k
=
1
k
{\displaystyle \delta _{k}={\frac {1}{k}}}
,
∃
A
k
⊂
X
{\displaystyle \exists A_{k}\subset X}
where
(
d
i
a
m
(
A
k
)
<
δ
k
)
{\displaystyle (diam(A_{k})<\delta _{k})}
and
¬
∃
β
(
A
k
⊂
U
β
)
{\displaystyle \neg \exists \beta (A_{k}\subset U_{\beta })}
⋮
For all
n
∈
Z
+
{\displaystyle n\in \mathbb {Z} ^{+}}
,
A
n
≠
∅
{\displaystyle A_{n}\neq \emptyset }
since
A
n
⊄
U
β
{\displaystyle A_{n}\not \subset U_{\beta }}
.
It is therefore possible to generate a sequence
{
x
n
}
{\displaystyle \{x_{n}\}}
where
x
n
∈
A
n
{\displaystyle x_{n}\in A_{n}}
by axiom of choice. By sequential compactness, there exists a subsequence
{
x
n
k
}
,
k
∈
Z
+
{\displaystyle \{x_{n_{k}}\},k\in \mathbb {Z} ^{+}}
that converges to
x
0
∈
X
{\displaystyle x_{0}\in X}
.
Using the fact that
A
{\displaystyle {\mathcal {A}}}
is an open covering,
∃
α
0
∈
J
{\displaystyle \exists \alpha _{0}\in J}
where
x
0
∈
U
α
0
{\displaystyle x_{0}\in U_{\alpha _{0}}}
. As
U
α
0
{\displaystyle U_{\alpha _{0}}}
is open,
∃
r
>
0
{\displaystyle \exists r>0}
such that
B
d
(
x
0
,
r
)
⊂
U
α
0
{\displaystyle B_{d}(x_{0},r)\subset U_{\alpha _{0}}}
. By definition of convergence,
∃
L
∈
Z
+
{\displaystyle \exists L\in \mathbb {Z} ^{+}}
such that
x
n
p
∈
B
d
(
x
0
,
r
2
)
{\displaystyle x_{n_{p}}\in B_{d}\left(x_{0},{\frac {r}{2}}\right)}
for all
p
≥
L
{\displaystyle p\geq L}
.
Furthermore,
∃
M
∈
Z
+
{\displaystyle \exists M\in \mathbb {Z} ^{+}}
where
δ
M
=
1
K
<
r
2
{\displaystyle \delta _{M}={\frac {1}{K}}<{\frac {r}{2}}}
. So,
∀
z
∈
Z
+
z
≥
M
⇒
d
i
a
m
(
A
M
)
<
r
2
{\displaystyle \forall z\in \mathbb {Z} ^{+}z\geq M\Rightarrow diam(A_{M})<{\frac {r}{2}}}
.
Finally, let
q
∈
Z
+
{\displaystyle q\in \mathbb {Z} ^{+}}
such that
n
q
≥
M
{\displaystyle n_{q}\geq M}
and
q
≥
L
{\displaystyle q\geq L}
. For all
x
′
∈
A
n
q
{\displaystyle x'\in A_{n_{q}}}
, notice that:
d
(
x
n
q
,
x
′
)
≤
d
i
a
m
(
A
n
q
)
<
r
2
{\displaystyle d(x_{n_{q}},x')\leq diam(A_{n_{q}})<{\frac {r}{2}}}
because
n
q
≥
M
{\displaystyle n_{q}\geq M}
.
d
(
x
n
q
,
x
0
)
<
r
2
{\displaystyle d(x_{n_{q}},x_{0})<{\frac {r}{2}}}
because
q
≥
L
{\displaystyle q\geq L}
which means
x
n
q
∈
B
d
(
x
0
,
r
2
)
{\displaystyle x_{n_{q}}\in B_{d}(x_{0},{\frac {r}{2}})}
.
By the triangle inequality,
d
(
x
0
,
x
′
)
<
r
{\displaystyle d(x_{0},x')<r}
, implying that
A
n
q
⊂
U
α
0
{\displaystyle A_{n_{q}}\subset U_{\alpha _{0}}}
which is a contradiction.