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Let $\{a_n\}$ be a bounded sequence. Prove that if every convergent subsequence of $\{a_n\}$ has limit $L$, then $\lim_{n\rightarrow\infty}a_n = L$.

I know that if the sequence has a limit, it must be $L$, because the limit must equal the limit of the subsequence. But how can I prove the sequence has a limit?

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    $\begingroup$ Suppose $(a_n)$ does not converge to $L$. What can you then say? $\endgroup$ – David Mitra Jun 9 '13 at 15:05
  • $\begingroup$ For some $\epsilon$, for any $N$ there exists $n>N$ such that $|a_n-L|>\epsilon$. So I can find a subsequence $b_1,b_2,\ldots$ such that $|b_i-L|>\epsilon$ for all $i$. But this is still not a contradiction, since $b_i$ might not be convergent... $\endgroup$ – Paul S. Jun 9 '13 at 15:17
  • $\begingroup$ Oh, but $b_i$ is a bounded sequence, so I can find a convergent subsequence $c_i$ of it (by Bolzano-Weierstrass). This subsequence $c_i$ cannot have limit $L$, a contradiction! $\endgroup$ – Paul S. Jun 9 '13 at 15:19
  • $\begingroup$ Please ignore my previous (now deleted) comment. $\endgroup$ – David Mitra Jun 9 '13 at 15:23
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Suppose $(a_i)$ does not converge to $L$. Then for some $\epsilon$, for any $N$ there exists $n>N$ such that $|a_n-L|>\epsilon$. So I can find a subsequence $b_1,b_2,\ldots$ such that $|b_i-L|>\epsilon$ for all $i$. Since $(b_i)$ is bounded, by Bolzano-Weierstrass I can find a convergent subsequence $(c_i)$ of it. Clearly $(c_i)$ cannot have limit $L$, a contradiction.

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Let $(x_n)$ be a sequence in a compact metric space $(M, d)$. Then the followings are equivalent:

  1. $(x_n)$ converges to $L$
  2. Every convergent subsequence of $(x_n)$ has limit $L$
  3. Every subsequence of $(x_n)$ has a further subseqeuence converging to $L$.

Proof.

  • $1 \Longrightarrow 2$ : Clear.
  • $2 \Longrightarrow 3$ : Let $(x_{n_k})$ be an arbitrary subsequence of $(x_n)$. Since $M$ is compact, it is sequentially compact and $(x_{n_k})$ has a convergent subsequence. Then its limit must be $L$ by the assumptions.
  • $3 \Longrightarrow 1$ : We prove the contraposition. Assume $x_{n}$ does not converge to $L$. Then there exists $\epsilon > 0$ such that infinitely many terms of $x_n$ satisfy $d(x_n, L) \geq \epsilon$. We collect such terms and enumerate then to obtain a subsequence $(x_{n_k})$. Then it is clear that no subsequence of $(x_{n_k})$ can converge to $L$.

We remark that the equivalence $1 \Longleftrightarrow 3$ holds for any metric space $M$. This can be used to prove that a certain topological is not metrizable.

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