Bolzano-Weierstrass theorem Every sequence $\{ x_n \}_{n=1}^\infty$ bounded in $\mathbb R$ has, at least, a convergent subsequence.

Proof Since $\{ x_n \}_{n=1}^\infty$ is bounded, then $\exists a,b\in \mathbb R$, $a< b$, so $\{ x_n \}_{n=1}^\infty \subset [a,b]$.

If $\{x_n\}_{n=1}^\infty$ is a finite set, then it has a constant subsequence, which is convergent.

If $\{x_n\}_{n=1}^\infty$ is an infinite set, because of the Bolzano-Weierstrass theorem, there exists an accumulation point $x_0$ of $\{x_n\}_{n=1}^\infty$ (already proved).

Let's build a subsequence of $\{x_n\}_{n=1}^\infty$ that converges to $x_0$: since $\forall \varepsilon > 0 \;\, B(x_0, \varepsilon)\setminus\{x_0\}\cap \{x_n\}_{n=1}^\infty \neq \emptyset$ (because of the definition of accumulation point), let $\varepsilon_1 =1$, then $\exists n_1 \in \mathbb{N} $ s.t. $ x_{n_1} \in B(x_0, \varepsilon_1)$.

Let $m_2 \in \mathbb{N}$, $\varepsilon_2 = \frac{1}{m_2} < d(x_{n_1}, x_0)$, then $\exists n_2\in \mathbb{N} \;\, (n_2 > n_1)$ s.t. $ x_{n_2} \in B(x_0, \varepsilon_2)$.

Inductively we have that $\{x_{n_k}\}_{k=1}^\infty$ is a subsequence of $\{x_n\}_{n=1}^\infty$ and $n_k < n_{k+1}$, so $ d(x_{n_k}, x_0) < \varepsilon_k = \frac{1}{m_k}$.

Then $\lim_{k\to\infty} d(x_{n_k}, x_0) = 0$, which is equivalent to $\lim_{k\to\infty} x_{n_k}= x_0$ so $\{x_{n_k}\}_{k=1}^\infty$ is a convergent subsequence.

Question How can I rewrite the last part (the inductive part seems unclear to me) and everything unclear, since I can't understand why $\forall \varepsilon > 0 \;\, \exists k_0 \;\, \forall k \geq k_0 \;\, d(x_{n_k}, x_0) < \varepsilon $ (def. of convergence)...? Why is built $\{m_k\}$?

  • $\begingroup$ The statement of the theorem is missing the word "bounded". $\endgroup$ – Asaf Karagila Dec 15 '12 at 17:15
  • $\begingroup$ Corrected, thanks. $\endgroup$ – user50554 Dec 15 '12 at 17:19
  • $\begingroup$ I don't think It is a good idea to write out the proof.This theorem has vivid geometric meaning. $\endgroup$ – Luqing Ye Dec 15 '12 at 17:22

The inductive part is unecessarily complicated. Here is a simpler version:

For $\epsilon=1$, $\exists n_1 \in \mathbb{N} $ s.t. $ x_{n_1} \in B(x_0, 1)$.

For $\epsilon=\frac12$, $\exists n_2\in \mathbb{N} \;\, (n_2 > n_1)$ s.t. $ x_{n_2} \in B(x_0, \frac12)$

Having defined $n_k$ so that $n_k>n_{k-1}$ and $ x_{n_k} \in B(x_0, \frac1k)$ define $n_{k+1}$ as follows:

For $\epsilon=\frac1{k+1}>0$, $\exists n_{k+1}\in \mathbb{N} \;\, (n_{k+1} > n_k)$ s.t. $ x_{n_{k+1}} \in B(x_0, \frac1{k+1})$

The subsequence $(x_{n_k})$ satisfies $ x_{n_k} \in B(x_0, \frac1k)$ or equivalently $$ d(x_{n_k}, x_0) <\frac{1}{k}$$ By the squeeze theorem, $$ d(x_{n_k}, x_0) \to 0$$ as $k\to +\infty$ and the proof is complete!

  • $\begingroup$ What about considering the constant subsequence? Should I remove that part? $\endgroup$ – user50554 Dec 15 '12 at 18:56
  • $\begingroup$ @user50554 You should. $\endgroup$ – Nameless Dec 15 '12 at 19:11

You are aiming to find a subsequence that converges to $x_0$. To do this, you are showing that given $\epsilon$>0, $\exists$ k$\epsilon$$\mathbb{N}$ such that B($x_0$;$\epsilon$) contains all terms in the subsequence that appear after the $k^{th}$ term. Now read the proof again with this in mind and let me know if you are still unable.

  • $\begingroup$ The idea isn't really the problem, the way of building the subsequence confuses me. $\endgroup$ – user50554 Dec 15 '12 at 17:43
  • $\begingroup$ You are taking natural numbers m_i such that m_i+1>m_i, and then taking e_i=1/m_i. Clearly m_i is decreasing. Consider the ball around x0 with radius e_i. x0 being accumulation point, this ball has at least one element of the sequence. Call this the ith term of your subsequence. $\endgroup$ – Sudhanshu Srivastava Dec 15 '12 at 17:56
  • $\begingroup$ $m_i$ is decreasing or $\varepsilon_i$? $\endgroup$ – user50554 Dec 15 '12 at 17:58
  • $\begingroup$ oh, so sorry. e_i is. and sorry for the bad typesetting. $\endgroup$ – Sudhanshu Srivastava Dec 15 '12 at 18:00

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