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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?

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2 Answers 2

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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, $s\leqslant t$.

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  • $\begingroup$ Your proof is more simple than mine. What's wrong with the proof by contadiction? $\endgroup$
    – RFZ
    Commented Aug 17, 2015 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
    Commented Oct 21, 2016 at 0:13
  • $\begingroup$ @AOrtiz One can assume that $(t_n)$ is bounded. $\endgroup$
    – Pedro
    Commented Oct 21, 2016 at 1:00
  • $\begingroup$ @PedroTamaroff Can you expand on why that is? $\endgroup$
    – Alex Ortiz
    Commented Oct 21, 2016 at 1:14
  • $\begingroup$ @AOrtiz Could you think about it for a while? I cannot do that now. $\endgroup$
    – Pedro
    Commented Oct 21, 2016 at 1:25
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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 $s_n$).

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