If $s_n\leqslant t_n$ for $n\geqslant N$, where $N$ is fixed, then $$\liminf_{n\to \infty} s_n\leqslant \liminf_{n\to \infty} t_n$$ $$\limsup_{n\to \infty} s_n\leqslant \limsup_{n\to \infty} t_n$$

Proof: I'll prove the second inequality (the proof is analogous in the other case). By Rudin's definition $\lim_{n\to \infty} \sup s_n=\sup S=s^*$ and $\limsup_{n\to \infty} t_n=\sup T=t^*$ where sets $E$ and $T$ contains all subsequential limits.

Let $\sup T=t^*\in \mathbb{R^1}$. First of all we'll prove that $\forall x\in S$ $\Rightarrow$ $x\leqslant \sup T=t^*$. If $x\in S$ then $\exists \{n_k\}:$ $\lim_{k\to \infty} s_{n_k}= x$ and $s_{n_k}\leqslant t_{n_k}$. Let by contradiction $\lim_{k\to \infty} s_{n_k}= x>t^*$. For $\varepsilon =\frac{x-t^*}{4}$ $\quad\exists N_{\varepsilon}: \forall n_k\geqslant N_{\varepsilon}$ $\Rightarrow$ $s_{n_k}\in(x-\varepsilon, x+\varepsilon)$.

Also $\forall n_k\geqslant N_{\varepsilon}$ $\Rightarrow$ $t_{n_k}>x-\varepsilon>t^*.$

If $t_{n_k}$ not bounded then exists some subsequence that tends to $+\infty$ and $+\infty\in E$ and we got contadiction.

If $t_{n_k}$ is bounded then by Bolzano–Weierstrass theorem exists subsequence $\{t_{n_{k_j}}\}$ s.t. $t_{n_{k_j}}\to t_1$ where $t_1\geqslant x-\varepsilon>t^*$ and $t_1\in E$ and we got that $t^*$ is not $\sup T$. Contradiction. We proved that $\forall x\in S$ $\Rightarrow$ $x\leqslant \sup T=t^*$ then $\sup S\leqslant \sup T$.

Cases when $\sup T=\pm \infty$ is obvious.

Is my proof correct?


It is not in your best interest to phrase every proof in terms of contradictions. Using the definitions you should be able to give a direct proof. By definition $s=\limsup\limits_{n\to\infty} s_n$ is the largest value of a convergent subsequence of $(s_n)$. Now for a fixed convergent subsequence $(s_{n_k})$ we get a subsequence $(t_{n_k})$ that might or might not converge. Pick a convergent subsequence $(t_{m_j})$, then $(s_{m_j})$ still converges to the same limit $s^*$ as $s_{n_k}$ but now $$s^*\leqslant \lim_{j\to \infty} t_{m_j}\leqslant t=\limsup_{n\to\infty} t_n$$ Since $s^*$ was an arbitrary subsequential limit, $s\leqslant t$.

  • $\begingroup$ Your proof is more simple than mine. What's wrong with the proof by contadiction? $\endgroup$ – ZFR Aug 17 '15 at 7:47
  • $\begingroup$ How do you know that $(s_{m_j})$ is a subsequence of $(s_{n_k})$? I see no reason why $(m_j)$ has to be a subsequence of $(n_k)$. Better, how do you know that there is a subsequence $(t_{m_j})$ of $(t_{n_k})$ such that $(t_{m_j})$ converges? $\endgroup$ – Alex Ortiz Oct 21 '16 at 0:13
  • $\begingroup$ @AOrtiz One can assume that $(t_n)$ is bounded. $\endgroup$ – Pedro Tamaroff Oct 21 '16 at 1:00
  • $\begingroup$ @PedroTamaroff Can you expand on why that is? $\endgroup$ – Alex Ortiz Oct 21 '16 at 1:14
  • $\begingroup$ @AOrtiz Could you think about it for a while? I cannot do that now. $\endgroup$ – Pedro Tamaroff Oct 21 '16 at 1:25

Here is a sketch of a solution using a previous theorem from Rudin's:

Theorem 3.17(b): If $s_X=\limsup{s_n}<x$ then $\exists N ~ | ~ n\ge N \implies s_n<x$

If $t_X < s_X$ then $\exists ~ x $ real such that $t_X < x < s_X$. By T3.17 and the hypothesis we have $s_n \le t_n < x$ for big enough $n$. But then no subsequence $s_{n_k} \rightarrow d $ for $d > x$ which contradicts T3.17a ($s^X$ is a subsequential limit of $s_n$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.