# Proof for sequences: $\limsup_{n \to \infty}(b_n - a_n) \le \limsup_{n \to \infty}{b_n} - \liminf_{n \to \infty}{a_n}$

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 $$\{b_n\}$$.

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.

• 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 $a_i$. Commented Nov 12, 2016 at 21:20
• @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.)
– Pedro
Commented Nov 12, 2016 at 21:42
• @Pedro Tamaroff♦ in your answer, $\{a_{k_{i}}\}$ is some convergent subsequence of $\{a_i\}$ , but from this how could you say that any one of lim sup or lim inf of $\{a_i\}$ exists? And notice that, inf of $\{b_{i}-a_{i}\}$ may not even exist. Commented Oct 23, 2020 at 7:28