- #1
evinda
Gold Member
MHB
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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)
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)
