Prove: $a_n \leq b_n \implies \limsup a_n \leq \limsup b_n$ [duplicate]

Is my proof correct? Prove: $a_n \leq b_n \implies \limsup a_n \leq \limsup b_n$

Proof:

Let $a_n$ and $b_n$ be sequences such that $a_n \leq b_n \forall_n$. Suppose $\limsup a_n \nleq \limsup b_n$.

That is: $\limsup a_n > \limsup b_n$. From this we know:

$\forall_{\epsilon > 0} \exists_N \forall_{n>N} \implies |b_n - b| <\epsilon$. Where $b$ is the $\limsup b_n$.

$\forall_{\epsilon_1 > 0} \exists_{N_1} \forall_{n > N_1} \implies |a_n - a| < \epsilon_1$. Where $a$ is the $\limsup a_n$

So, let $a^* = a + \dfrac{\epsilon_1}{2}$ and let $b^* = b - \dfrac{\epsilon}{2}$.

Hence, $a^* \in |a_n - a| <\epsilon_1$ and $b^* \in |b_n - b| < \epsilon$. And clearly we see that $b^* < a^*$.Thus, we have found an element of $b_n$ namely $b^* < a^*$ an element of $a$. This is contradiction since we are given $a_n \leq b_n \forall_n$.

Therefore, the supposition is false, and $\limsup a_n \leq \limsup b_n$.

Solution 1:

Recall that given a bounded sequence, we define $\limsup a_n$ as $$\lim a_n^+$$ where $$a_n^+=\sup\{a_n,a_{n+1},\dots,\}$$

Now, if $a_n\leq b_n$ for each $n$, what can you say about the relaton among each of the values:

$$a_n^+=\sup\{a_n,a_{n+1},\dots,\}$$ $$b_n^+=\sup\{b_n,b_{n+1},\dots,\}$$

What can you then say about their limits?