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Bolzano-Weierstrass Theorem 📂Analysis

Bolzano-Weierstrass Theorem

Theorem

If a given infinite set ERE \subset \mathbb{R} is bounded, then there is an accumulation point pRp \in \mathbb{R} of EE.

Explanation

Or one could also say, ‘A bounded sequence has a converging subsequence.’ It’s important to note that EE does not need to be closed under the condition.

Proof

Part 1. n=1In={x}\displaystyle \bigcap_{n=1}^{\infty} I_{n} = \left\{ x \right\}

Since EE is bounded by assumption, there exists a closed interval I1:=[a,b]I_{1} := [a,b] that satisfies EI1E \subset I_{1}. Here, let’s take I2I_{2} as an interval that has half the length of I1I_{1} and contains infinitely many points of EE. By repeating this process, we can create a nested interval In+1InI_{n+1} \subset I_{n}, and the length of InI_{n} becomes dn:=ba2n\displaystyle d_{n} : = {{b-a} \over {2^{n}}}.

Cantor’s Interval Theorem: For nested intervals [an,bn][a_{n}, b_{n}], if limn(bnan)=0\displaystyle \lim_{n \to \infty} (b_{n} - a_{n}) = 0 then n=1[an,bn]\displaystyle \bigcap_{n=1}^{\infty} [a_{n}, b_{n}] is a singleton set.

Since limndn=0\displaystyle \lim_{n \to \infty} d_{n} = 0, by Cantor’s Interval Theorem, n=1In={x}\displaystyle \bigcap_{n=1}^{\infty} I_{n} = \left\{ x \right\} [NOTE : xEx \in E is not necessarily required]


Part 2. xx is an accumulation point of EE

Considering an open set OO that contains xx, due to the density of real numbers, there exists a ε>0\varepsilon>0 that makes (xε,x+ε)O(x - \varepsilon , x + \varepsilon ) \subset O. As previously shown, since limndn=0\displaystyle \lim_{n \to \infty} d_{n} = 0, a natural number n0n_{0} that satisfies dn0=ba2n0<ε\displaystyle d_{n_{0}} = {{b-a} \over {2^{n_{0}}}} < \varepsilon will also exist so

xIn0(xε,x+ε)O x \in I_{n_{0}} \subset (x - \varepsilon , x + \varepsilon ) \subset O

Accumulation point: For a point xRx \in \mathbb{R} in real numbers and a subset ARA \subset \mathbb{R}, if for any open set OO containing xx, O(A{x}) O \cap ( A \setminus \left\{ x \right\} ) \ne \emptyset then xx is defined as an accumulation point.

No matter how OO is defined, as long as it is an open set containing xx, (O{x})E\left( O \setminus \left\{ x \right\} \right) \cap E \ne \emptyset is inevitable. Therefore, xx becomes an accumulation point of EE.

See Also