Real analysis
Let A,A1,A2,…∈℘(X). A1⊆A2⊆… and ⋃n∈NAn=A⟺ the sequence (An)n∈N increases to A.
Similarly, A1⊇A2⊇… and ∩n∈NAn=A⟺ the sequence (An)n∈N decreases to A.
∀x∈X,limn→∞1An(x)=1A(x)⟺(An)n∈N converges to A. limn→∞An=A
- limsupn→∞An:=∩n∈N⋃k≥nAk and liminfn→∞An:=⋃n∈N∩k≥nAk
- Since ∀l,n∈N,∩k≥lAk⊆⋃k≥nAk, then liminfn→∞An⊂limsupn→∞An
Binomial Formula: (a+b)n=k=0∑nnCk⋅akbn−k
Multinomial Formula: (a1+a2+⋯+al)n=∑k1,k2,…,klk1!k2!…kl!n!∏i=1laiki
The indicator function on A∈℘(X), denote as 1A:X→R where 1A(x):={10x∈Ax∈/A
- 1A∩B=1A1B
- A∩B=∅⟹1A∪B=1A+1B
- 1A+1Ac=1