Difference between revisions of "Theorem:Bolzano-Weierstrass"
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Now we show there is a subsequence that converges to ''x''. We define the subsequence inductively, choose any ''x''<sub>n<sub>1</sub></sub> from the interval (''x'' − 1, ''x'' + 1). Assuming we have chosen ''x''<sub>''n''<sub>1</sub></sub>, …, ''x''<sub>''n''<sub>''k''−1</sub></sub>, choose ''x''<sub>''n''<sub>''k''</sub></sub> to be an element in the interval (''x'' − 1/k, ''x'' + 1/k) so that ''n''<sub>''k''</sub>∉{''n''<sub>1</sub>, …, ''n''<sub>k−1</sub>}, this is possible as there are infinitely many elements of (''x''<sub>''n''</sub>) in the interval. Notice that for this choice of ''x''<sub>''n''<sub>''k''</sub></sub> we have that |''x'' − ''x''<sub>''n''<sub>''k''</sub></sub>|<1/k. Hence for any ε>0, if we take any ''k'' > 1/ε, then |''x''<sub>''n''<sub>''k''</sub></sub>-x| < ε. That is the subsequence (''x''<sub>''n''<sub>k</sub></sub>) → ''x''. <math>\Box</math> | Now we show there is a subsequence that converges to ''x''. We define the subsequence inductively, choose any ''x''<sub>n<sub>1</sub></sub> from the interval (''x'' − 1, ''x'' + 1). Assuming we have chosen ''x''<sub>''n''<sub>1</sub></sub>, …, ''x''<sub>''n''<sub>''k''−1</sub></sub>, choose ''x''<sub>''n''<sub>''k''</sub></sub> to be an element in the interval (''x'' − 1/k, ''x'' + 1/k) so that ''n''<sub>''k''</sub>∉{''n''<sub>1</sub>, …, ''n''<sub>k−1</sub>}, this is possible as there are infinitely many elements of (''x''<sub>''n''</sub>) in the interval. Notice that for this choice of ''x''<sub>''n''<sub>''k''</sub></sub> we have that |''x'' − ''x''<sub>''n''<sub>''k''</sub></sub>|<1/k. Hence for any ε>0, if we take any ''k'' > 1/ε, then |''x''<sub>''n''<sub>''k''</sub></sub>-x| < ε. That is the subsequence (''x''<sub>''n''<sub>k</sub></sub>) → ''x''. <math>\Box</math> | ||
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| + | ==Resources== | ||
| + | * [https://en.wikibooks.org/wiki/Real_Analysis/Sequences Sequences], Wikibooks: Real Analysis | ||
Revision as of 15:12, 14 October 2021
Theorem (Bolzano—Weierstrass)
Every bounded sequence of real numbers contains a convergent subsequence.
Let (xn) be a sequence of real numbers bounded by a real number M, that is |xn| < M for all n. We define the set A by A = {r | |r| ≤ M and r < xn for infinitely many n}. We note that A is non-empty since it contains −M and A is bounded above by M. Let x = sup A.
We claim that, for any ε > 0, there must be infinitely many points of xn in the interval (x − ε, x + ε). Suppose not and fix an ε > 0 so that there are only finitely many values of xn in the interval (x − ε, x + ε). Either x ≤ xn for infinitely many n or x ≤ xn for at most only finitely many n (possibly no n at all). Suppose x< xn for infinitely many n. Clearly in this case x ≠ M. If necessary restrict ε so that x + ε ≤ M. Set r = x + ε/2 we have that r < xn for infinitely many n because there are only finitely many xn in the set [x,r] and x must be less than infinitely many xn, furthermore |r| < M. Thus r is in A, which contradicts that x is an upper bound for A. Now suppose x< xn for at most finitely many n. Set y = x − ε/2. Then there are at most only finitely man n so that xn ≥ y. Thus, if r < xn for infinitely many n, we have that r ≤ y. This means that y is an upper bound for A that is less than x, contradicting that x wast the least upper bound of A. In either case we arrive at a contradiction, thus we must have that for any ε > 0, there must be infinitely many points of xn in the interval (x − ε, x + ε).
Now we show there is a subsequence that converges to x. We define the subsequence inductively, choose any xn1 from the interval (x − 1, x + 1). Assuming we have chosen xn1, …, xnk−1, choose xnk to be an element in the interval (x − 1/k, x + 1/k) so that nk∉{n1, …, nk−1}, this is possible as there are infinitely many elements of (xn) in the interval. Notice that for this choice of xnk we have that |x − xnk|<1/k. Hence for any ε>0, if we take any k > 1/ε, then |xnk-x| < ε. That is the subsequence (xnk) → x.
Resources
- Sequences, Wikibooks: Real Analysis
