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.

  • $\begingroup$ 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

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.


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