Compactness of a Set
Course Title : Real Analysis 1
Course Code : MTH-211
Submitted to
Submitted by
Abdullah Al Mahbub
Maruf Ahmed Abeer
Associate Professor
Id: 12004025
Department of Mathematics
Session: 2019-20
Comilla University
Department of Mathematics
Comilla University
Sneak peek
▪ Open cover of a set
▪ Subcover of a set
▪ Compact Set
Compactness
Open Cover of Set
Let 𝑆⊂ R and 𝐶 = {𝐴 |𝑖 ∈ 𝐼} be a collection of open sets then C is
said to be open cover of S if
𝑆
𝑖" ∈" 𝐼 𝐴𝑖
Example : Let S = R (Real numbers) and
C = {(-n, n) | n ∈ N} then 𝑺 ⊂
(
…..-4
(
-3
(
-2
(
-1
)
1
𝒏∈𝑵 (−𝒏, 𝒏)
)
2
)
3
)
4 …….
C = {(-n, n) | n ∈ N} is an open cover of R
Subcover:
A sub-collection of open cover of any set is called a
subcover.
Example : C = {(-n, n) | n ∈ N} and D = {(-3n, 3n) |
n<4 & n ∈ N}
Here both are open cover of R but D is subcover
of C
Compact Set:
If a finite subcover of S is contained in each open cover of any set S,
then S is called a compact set.
Example : A finite subset of real number is compact.
Proof Let S = {a₁, a,....... a} is a finite set and G be the family of sub
cover of S.
|
|
a1
a2
|
|
a3… … … … … .. an
Then ∃ a open set containing each a ∈ S
(| )
(| )
(| )
( | )
a1.
a2
a3. ….. … … an
Hence S is finite → open sets containing a are finite.
→G has finite subcover of S
⇒ S is compact.
Example – 2:
R = {set of real number} is not compact.
Proof: We know that R has a open cover C = {(-n, n): n ∈ N} and
'C' has infinite subcover like ((-kn, kn): k N, n ∈N}
→ R is not compact.
The End