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Let $(u_n)_{n \in \mathbb{N}}$ be an unbounded above sequence. Prove that it has a subsequence which diverges to $+\infty$

We have : $\forall k \in \mathbb{R},\exists n \in \mathbb{N};u_n>k$

So let $\phi$ such that $\phi(0)=n_0$ (for $k=0$) and $\phi(n+1)=\inf(p \in \mathbb{N};u_p>\max(u_{\phi(n)},n+1))$

then $u_{\phi(n)}>n$ and $\phi(n+1) \neq \phi(n).$

How can we prove that $\phi(n+1)>\phi(n)?$

postmortes
  • 6,338
  • My suggestion, don't. Define $$\phi(n+1)=\inf{p>\phi(n):u_{p}>\max(u_{\phi(n)},n+1)}.$$ – Floris Claassens Jun 24 '19 at 12:22
  • @FlorisClaassens why can you define it like this? –  Jun 24 '19 at 13:04
  • is it because a sequence is bounded above if and only if $\exists l \in \mathbb{N}$ such that $(u_n)_{n \geq l}$ is bounded? –  Jun 24 '19 at 13:08
  • If a sequence is unbounded, then the set $X_{k}={n:u_{n}>k}$ is of infinite cardinality. If this was not true, then let $X_{k}$ has a maximum and $$X_{\max X_{k}}={n:u_{n}>\max X_{k}}=\emptyset,$$ which contradicts with the sequence being unbounded. – Floris Claassens Jun 24 '19 at 14:27

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