# If $a_n\leq b_n$, then $\lim \sup a_n \leq \lim b_n$ [duplicate]

Hello I want to make sure my work is correct!

Suppose that $\{a_n\}$ and $\{b_n\}$ are sequences such that $a_n \leq b_n$ for all $n$ and $b_n \to b$. Prove that $\lim \sup a_n \leq b.$

Let $\epsilon > 0$ be given. Choose an $N$ s.t. $|b_n-b|<\epsilon$ for $n\geq N$ where $b = \lim \sup b_n$.

$a = \lim \sup a_n$.

Proof by contradiction: If $a \geq b$, then, since $a_n\leq b$,

$b\leq a\leq b_n$, meaning $|b_n-a|<|b_n-b|<\epsilon$, implying $b_n \to a$. This is a contradiction,

therefore, $a\leq b$.

## marked as duplicate by Pedro Tamaroff♦, Sujaan Kunalan, Dominic Michaelis, Old John, azimutNov 18 '13 at 8:37

No. It may happen that $b_n \to a$. That is not a problem.
What you can do is as follows : You know that $a_n \leq b_n$ for all $n \in \mathbb{N}$. Hence, for any $k \in \mathbb{N}$ $$u_k := \sup\{ a_n : n\geq k \} \leq \sup\{ b_n : n\geq k\} =: v_k$$ Now, $v_k \to b$, and hence $$\limsup a_n = \lim u_k \leq \lim v_k = b$$
Another approach: Let $c_n:=b_n-a_n \geq 0$ . Since $c_n \geq 0$, $LimSup c_n >0$.
Claim: Let $c_n$ be a sequence of nonnegative terms. Then $LimSup c_n>0$.
Proof of Claim: $LimSup${$c_n$}$:=c$ is , by definition, the largest limit point of $c_n$ , so that, in particular, it is a limit point of the sequencee $c_n$. Assume, BWC , that $C_n$ has $c<0$ as a limit point. But then the ball $B(c, |c|/2)=(c-|c|/2,c+|c|/2)$ will not intersect {$c_n$} , since $c+|c|/2=c-c/2=c/2<0$ , i.e., the neighborhood $B(c, |c|/2)$ of $c$ that does not intersect the sequence, so that $c<0$ cannot be a limit point of {$c_n$}. It follows that $LimSup$ {$c_n$}$:=b_n-a_n \geq 0$, so that $LimSup a_n \geq LimSup$b_n\$