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If $\{a_n\}$ and $\{b_n\}$ are real-valued sequences, and $\{b_n\}$ is bounded, prove that: $\limsup_{n \to \infty}(b_n – a_n) \le \limsup_{n \to \infty}{b_n} - \liminf_{n \to \infty}{a_n}$ .

Please, check whether this proof is correct:

Define $B_k = \sup\{b_n: n \ge k\}$. So $\limsup {b_n} = \lim {B_k}$.

Define $A_k = \inf\{a_n: n \ge k\}$. So $-A_k = \sup\{-a_n: n \ge k\}$ .

Define $C_k = \sup\{b_n + (- a_n): n \ge k\}$.

Fix k. Then $b_n +(– a_n) \le B_k +(– A_k)$.

We know that $\sup(X + Y) = \sup(X) + \sup(Y)$.

So $C_k = \sup{(b_n – a_n)} \le B_k – A_k$. (But why does this inequality follow from previous equality?).

Therefore $\limsup{(b_n – a_n)} \le \limsup{b_n} - \liminf{a_n}$.

The last equality follows from the fact that $\sup(-a_n) = -\inf(a_n)$ and taking limits on the both sides of the inequality, which preserves the inequality.

Another thing I am unsure about is where I should use boundedness of $\{b_n\}$.

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Suppose that $(b_{k_i}-a_{k_i})$ is a convergent subsequence of $(b_i-a_i)$. Because $(b_i)$ is bounded (use Bolzano-Weiertrass), there is a further subsequence --- for the sake of notation, keep the same indices --- $(b_{k_i}-a_{k_i})$ such that $b_{k_i}$ converges. It follows that $a_{k_i} = b_{k_i}-(b_{k_i}-a_{k_i})$ converges too. By definition of the limit superior and limit inferior (they are the largest and smallest subsequential limit of a sequence), we have that

$$\begin{align*} \lim_{i\to\infty} (b_{k_i}-a_{k_i}) &= \lim_{i\to\infty} b_{k_i}- \lim_{i\to\infty} a_{k_i} \\ &\leqslant \limsup b_k + \limsup(-a_k) \\ &\leqslant \limsup b_k - \liminf a_k \end{align*}$$

Because we can now choose $(b_{k_i}-a_{k_i})$ so that

$$\lim_{i\to\infty} (b_{k_i}-a_{k_i}) = \limsup(b_k-a_k)$$

we conclude.

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  • $\begingroup$ But how do we know that convergent subsequence $(b_{k_i}-a_{k_i})$ exists? $b_i$ has a convergent subsequence by Bolzano-Weierstrass, but we do not know anything about $a_i$. $\endgroup$ – User 1234 Nov 12 '16 at 21:20
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    $\begingroup$ @MoyseyAbramowitz I never claimed one such subsequence existed (but they do, in the extended sense). I'm picking one, because $\limsup (b_k-a_k)$ is the largest limit of a convergent subsequence (which might possibly be $-\infty$ or $+\infty$. In this case the claim is evident.) $\endgroup$ – Pedro Tamaroff Nov 12 '16 at 21:42

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