Sequence has convergent subsequence

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SUMMARY

The discussion centers on the convergence of a bounded sequence $(b_n)$ that lies between two other bounded sequences $(a_n)$ and $(c_n)$. It establishes that since both $(a_n)$ and $(c_n)$ are bounded, the sequence $(b_n)$ must also be bounded. By applying the Bolzano-Weierstrass theorem, it is concluded that $(b_n)$ possesses a convergent subsequence. The participants confirm the correctness of this reasoning, affirming the completeness of the argument.

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evinda
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Hello! (Wave)

Let $(a_n), (b_n), (c_n)$ sequences such that $(a_n), (c_n)$ are bounded and $a_n \leq b_n \leq c_n$ for each $n=1,2, \dots$ I want to show that $(b_n)$ has a convergent subsequence.

I have thought the following:

Since $(a_n), (c_n)$ are bounded, $\exists m_1, m_2 \in \mathbb{Z}$ such that $m_1 \leq a_n$ and $c_n \leq m_2$.

Then $m_1 \leq b_n \leq m_2$, i.e. $(b_n)$ is bounded. So, from Bolzano-Weierstrass theorem, $(b_n)$ has a convergent subsequence.

But... is the above complete? Could we also show it somehow else? (Thinking)
 
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Hey evinda,

Looks good to me.
 
GJA said:
Hey evinda,

Looks good to me.

Nice, thank you... (Smirk)
 

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