0
$\begingroup$

Proof:

For any $N \in \mathbb{N}$, sup $(T_n) \geq x_n$ for all $n \geq N$.

Similarly, for any $N \in \mathbb{N}$, inf $(T_n) \leq x_n$ for all $n > \geq N$.

Hence, sup $(T_n) \geq$ inf $(T_n)$. Since this is true for any $N \in > \mathbb{N}$, lim sup $(x_n) \geq$ lim inf $(x_n)$.

Although I am convinced that my proof is correct, I feel like I am not giving enough mathematical rigours. Actually, I have this problem for all the proof problems I solve.

$\endgroup$
  • $\begingroup$ You can't do better. $\endgroup$ – Guru Jan 9 '17 at 19:25
  • $\begingroup$ Thank you for the comment! $\endgroup$ – user3000482 Jan 9 '17 at 19:25
  • $\begingroup$ In the first sentence of your proof, one can guess though, what is $T_n$? $\endgroup$ – Jack Jan 9 '17 at 20:11
1
$\begingroup$

Your method is solid. Sounds like you've reduced the problem to: "if $a_n\ge b_n $ $\forall n > N,$ (and a and b are monotone/convergent in $[-\infty,\infty]$) then $\lim_{n\rightarrow\infty} a_n \ge\lim_{n\rightarrow\infty} b_n$" and are uncomfortable with not giving a rigorous proof of this very plausible fact.

To prove this, let $L_a$ and $L_b$ denote the (finite) limits, and assume the contrary that $L_b>L_a + \epsilon$ for some $\epsilon>0.$ observe that there is an $N$ such that $|a_n-L_a|<\epsilon/2$ and $|b_n-L_b|<\epsilon/2$ for all $n>N$ and thus (draw a picture) $b_n>a_n$ for $n>N$. Then handle the cases where one or both limits is infinite.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.