Bolzano-Weierstrass Theorem
📂AnalysisBolzano-Weierstrass Theorem
Theorem
If a given infinite set E⊂R is bounded, then there is an accumulation point p∈R of E.
Explanation
Or one could also say, ‘A bounded sequence has a converging subsequence.’ It’s important to note that E does not need to be closed under the condition.
Proof
Part 1. n=1⋂∞In={x}
Since E is bounded by assumption, there exists a closed interval I1:=[a,b] that satisfies E⊂I1. Here, let’s take I2 as an interval that has half the length of I1 and contains infinitely many points of E. By repeating this process, we can create a nested interval In+1⊂In, and the length of In becomes dn:=2nb−a.
Cantor’s Interval Theorem: For nested intervals [an,bn], if n→∞lim(bn−an)=0 then n=1⋂∞[an,bn] is a singleton set.
Since n→∞limdn=0, by Cantor’s Interval Theorem, n=1⋂∞In={x} [NOTE : x∈E is not necessarily required]
Part 2. x is an accumulation point of E
Considering an open set O that contains x, due to the density of real numbers, there exists a ε>0 that makes (x−ε,x+ε)⊂O. As previously shown, since n→∞limdn=0, a natural number n0 that satisfies dn0=2n0b−a<ε will also exist so
x∈In0⊂(x−ε,x+ε)⊂O
Accumulation point: For a point x∈R in real numbers and a subset A⊂R, if for any open set O containing x, O∩(A∖{x})=∅ then x is defined as an accumulation point.
No matter how O is defined, as long as it is an open set containing x, (O∖{x})∩E=∅ is inevitable. Therefore, x becomes an accumulation point of E.
■
See Also