Prove if $\forall n\:a_n\ge b_n$ then $\liminf _{n\to \infty }\left(b_n\right)\le \limsup _{n\to \infty }\left(a_n\right)$

my suggest prove :

Let $(b_{n_k})_k$ subsequence of $(b_n)_n$ that convergence to $limsup \ b_n $

$(a_{n_k})_k$ is bounded because $(a_n)_n$ is bounded so does $(a_{n_k})_k$ then (by Bolzano–Weierstrass) $(a_{n_k})_k$ has convergence subsequence $(a_{n_{k_l}})_l$

so for every l : $$b_{n_{k_l}} \leq a_{n_{k_l}} $$

because for $(b_{n_{k_l}})_l$ is subsequence of $(b_{n_k})_k$ and convergence to $\limsup b_n $ then

$$\limsup b_n \leq \lim a_{n_{k_l}}$$

and because $a_n$ there is subsequence limit that great and equal $\limsup b_n$ then $\limsup b_n \leq \limsup a_n$

and becouse $\liminf b_n \leq \limsup b_n$ then $\liminf b_n \leq \limsup a_n$

is that prove correct and does i need to prove that $\liminf b_n \leq \limsup b_n$ ?

  • 1
    $\begingroup$ You did not state that $a_n$ is bounded in the problem? $\endgroup$ – Shashi Dec 7 '17 at 17:27
  • $\begingroup$ No, $\limsup$ can be infinity.. $\endgroup$ – Shashi Dec 7 '17 at 17:31
  • $\begingroup$ Ow, In my course we defined $\limsup$ to be infinity if the sequence was unbounded from above. Everyone his definitions.. But how did you define it @E.Roi ? What is the definition you use? $\endgroup$ – Shashi Dec 7 '17 at 17:32
  • $\begingroup$ hmm.. so i think this prove are not true unless the sequences are bounded ? $\endgroup$ – E.Roi Dec 7 '17 at 17:38
  • $\begingroup$ I have added an answer with two possible definitions for $\limsup$ and $\liminf$. $\endgroup$ – Shashi Dec 7 '17 at 17:52

If you use the following definition $$\limsup a_n := \inf_{i\in\mathbb{N}} \ \sup\{ a_k : k\geq i \} $$ and $$\liminf b_n := \sup_{i\in\mathbb{N}} \ \inf\{ b_k : k\geq i \}$$

Then we surely have $\sup\{ a_k : k\geq i \} \geq \inf\{ b_k : k\geq i \}$, so $\liminf b_n \leq \limsup a_n$ by the definition of supremum and infimum.

If you use the definition that says $\limsup$ is the largest value of the limit of the subsequences and $\liminf$ is the smallest value of the limit of the subsequences, then we must do something else. Let $b_{n_k}$ be the subsequence that converges to $\liminf b_n$ and $a_{n_j}$ the subsequence that converges to $\limsup a_n$.

Assume now that $\lim_{k\to \infty} b_{n_k} > \lim_{j\to \infty} a_{n_j}$. Now consider the subsequence $a_{n_k}$ that have the same index as $b_{n_k}$. We clearly have $\lim_{k\to \infty} a_{n_k} > \lim_{j\to \infty} a_{n_j}$. But we just said that $a_{n_j}$ was the sequence that was convergent to the highest value possible. Hence a contradiction.


Whether or not $(a_{n})$ is bounded, we always can realise $\limsup_{a}a_{n}$ to a subsequence $(a_{n_{k}})$ of $(a_{n})$: $\limsup_{n}a_{n}=\lim_{k}a_{n_{k}}$, of course in the sense of extended real, which means, it could be the case that $\lim_{k}a_{n_{k}}=\infty$ or $-\infty$. Similar argument applied to $\liminf_{n}b_{n}$

Now we know that $\limsup_{n}a_{n}$ is the greatest limit points of all its subsequences, and $\liminf_{n}b_{n}$ is the least. Using this, realising $\liminf_{n}a_{n}=\lim_{k}a_{n_{k}}$, for this subsequence $(a_{n_{k}})$, we have $b_{n_{k}}\leq a_{n_{k}}$. Realising $\liminf_{k}b_{n_{k}}=\lim_{l}b_{n_{k_{l}}}$, and the subsequence $(b_{n_{k_{l}}})$ must satisfy $b_{n_{k_{l}}}\leq a_{n_{k_{l}}}$. However, we have $\lim_{l}a_{n_{k_{l}}}=\lim_{k}a_{n_{k}}$, $\lim_{l}b_{n_{k_{l}}}\leq\lim_{l}a_{n_{k_{l}}}$, and $\liminf_{n}b_{n}\leq\lim_{l}b_{n_{k_{l}}}$, we actually prove that $\liminf_{n}b_{n}\leq\liminf_{n}a_{n}$.


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