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Consider the sequences $x_n$,$y_n\in\mathbb R$. We know that $x_n$ converges, but absolutely nothing is mentioned about $y_n$. I saw in a proof the following claim and I don't get the reasoning : $$ \lim \inf (x_n - y_n ) = \lim \inf (x_n ) + \lim \inf ( - y_n ) $$

Is this correct ?

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    $\begingroup$ Show the proof maybe? $\endgroup$ – BCLC Oct 15 '14 at 22:03
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    $\begingroup$ Why the downvote ? There's no need for any additional context since I've provided my question in the most general sense, my only condition is that $x_n$ converges . $\endgroup$ – Victor Oct 15 '14 at 22:09
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By definition, $$ \limsup_{n\to\infty}x_n=\lim_{k\to\infty}\sup_{n\ge k}x_n\tag{1} $$ and $$ \liminf_{n\to\infty}x_n=\lim_{k\to\infty}\inf_{n\ge k}x_n\tag{2} $$ Note that the $\inf$ of a sum is no less than the sum of the $\inf$s $$ \inf_{n\ge k}(x_n+y_n)=\overbrace{\inf_{\substack{m,n\ge k\\m=n}}(x_m+y_n)}^{\inf\text{ over a smaller set}}\ge\overbrace{\inf_{m,n\ge k}(x_m+y_n)}^{\inf\text{ over a larger set}}=\inf_{n\ge k}x_n+\inf_{n\ge k}y_n\tag{3} $$ Since $y_n=(x_n+y_n)+(-x_n)$, we can apply $(3)$: $$ \inf_{n\ge k}y_n\ge\inf_{n\ge k}(x_n+y_n)+\inf_{n\ge k}(-x_n)\tag{4} $$ Since $\sup\limits_{n\ge k}x_n=-\inf\limits_{n\ge k}(-x_n)$, $(4)$ becomes $$ \sup_{n\ge k}x_n+\inf_{n\ge k}y_n\ge\inf_{n\ge k}(x_n+y_n)\tag{5} $$ Combine $(3)$ and $(5)$: $$ \sup_{n\ge k}x_n+\inf_{n\ge k}y_n\ge\inf_{n\ge k}(x_n+y_n)\ge\inf_{n\ge k}x_n+\inf_{n\ge k}y_n\tag{6} $$ Taking the limit of $(6)$ yields $$ \lim_{n\to\infty}x_n+\liminf_{n\to\infty}y_n\ge\liminf_{n\to\infty}(x_n+y_n)\ge\lim_{n\to\infty}x_n+\liminf_{n\to\infty}y_n\tag{7} $$ Therefore, $$ \lim_{n\to\infty}x_n+\liminf_{n\to\infty}y_n=\liminf_{n\to\infty}(x_n+y_n)\tag{8} $$ Now just substitute $y_n\mapsto-y_n$.

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    $\begingroup$ Sorry I haven't marked an answer in such a long time, I completely forgot about this question.My sincere apologies @robjohn and thank you very much for your detailed proof ! $\endgroup$ – Victor Oct 24 '14 at 23:02
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    $\begingroup$ No problem. Voting is so unpredictable, I just answer questions as best as I can and the votes and acceptances are just happy occurrences sprinkled on later. $\endgroup$ – robjohn Oct 24 '14 at 23:49
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Indeed it is correct, and it can be simplified as $$ \liminf (x_n-y_n)=\lim x_n-\limsup y_n. $$

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    $\begingroup$ but he says $x_n$ converges... $\endgroup$ – Troy Woo Oct 15 '14 at 22:06
  • $\begingroup$ Yep, that's right.. @TroyWoo $\endgroup$ – Victor Oct 15 '14 at 22:09
  • $\begingroup$ I can't believe you accepted this. @Victor $\endgroup$ – Troy Woo Oct 15 '14 at 22:25
  • $\begingroup$ @TroyWoo I didn't, probably just misclicked. I always thank anyone who provides me with insightful answers (obviously not the case here). Thanks for the nudge $\endgroup$ – Victor Oct 15 '14 at 22:47
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Make use of the definition of $\liminf$ as the smallest limit of a convergent subsequence.

Since $x_n$ converges, if a subsequence of $(-y_n)$ converges then so does the corresponding subsequence of $(x_n-y_n)$. Moreover the limit of every such subsequence of $(x_n-y_n)$ is just the limit of the corresponding subsequence of $(-y_n)$ shifted by an amount equal to the limit of $x_n$ (since each subsequence of $x_n$ has this limit).

Since the amount of this shift does not vary from subsequence to subsequence, the subsequence for which the limit of $(-y_n)$ is smallest will also be the subsequence for which the limit of $(x_n-y_n)$ is the smallest and so we have

$$\liminf(x_n-y_n) = \lim x_n + \liminf (-y_n)$$

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