I have questions about the solution below.

I couldn't understand the red lines. What is $X_nN_1$? I'm not sure how it led to the contradiction.

Thank you!

Exercise 3: For any two real sequences $(a_n)$ and $(b_n)$, prove that $$\limsup_{n\to\infty} (a_n + b_n) \le \limsup_{n\to\infty} a_n + \limsup_{n\to\infty} b_n,$$ provided the sum on the right is not of the form $\infty-\infty$.

Proof. Assume that $\limsup_{n\to\infty}a_n=L<\infty$ (The proof with $\limsup_{n\to\infty}b_n$ finite works similarly).

Now, we break into three cases depending on whether $\limsup_{n\to\infty} b_n$ is finite, $\infty$, or $-\infty$.

If it's $-\infty$, we claim that both $\limsup_{n\to\infty}(a_n+b_n)$ and $\limsup_{n\to\infty}a_n + \limsup_{n\to\infty} b_n$ are $-\infty$. Note that it's clear that $$\limsup_{n\to\infty} a_n + \limsup_{n\to\infty} b_n = L-\infty = -\infty,$$ so we'll just show that $$\limsup_{n\to\infty}(a_n+b_n) = -\infty.$$ To see this, let $r$ be any real number. We must show $a_n+b_n<r$ for all sufficiently large $n$, for then it follows that any subsequence of $(a_n+b_n)$ eventually gets below $r$, and hence $\limsup_{n\to\infty}(a_n+b_n)\le r$. Doing this for all $r\in\Bbb R$ then shows that $\limsup_{n\to\infty}(a_n+b_n) = -\infty$.

To see that $a_n+b_n<r$ for sufficiently large $n$, notice first that since $\limsup_{n\to\infty}b_n=-\infty$, there must be an $N_1$ such that for all $n\ge N_1$, $b_n<r-|L|$. This is because otherwise, for each $N_1\in \Bbb N$, we could find an $x_{n_{N_1}}$ with $x_{n_{N_1}}\ge r$. Then the subsequence $(x_{n_{N_1}})$ would show that $\limsup_{n\to\infty}b_n \ge r$, a contradiction.

There must also be an $N_2$ such that for any $n \ge N_2$, $a_n < L$, since otherwise we'd find $L$ wasn't the limit superior of $(a_n)$. If we let $N = \max \{ N_1, N_2 \}$, then for any $n > N$, we have $a_n + b_n < L+ r-|L|\le r$. Thus, the claim is established and it follows that $\limsup_{n\to\infty}(a_n+b_n) = -\infty$.

Note: the original text can be found here: http://i.stack.imgur.com/sdkiM.png


Well, that's just a shoddy proof. Don't trust every PDF you find on the Internet!

  • The author writes $x_{n_{N_1}}$ but they meant $b_{n_{N_1}}$. I don't know where the letter "x" came from. And do we need the triple subscripts?
  • Every instance of $\star < r$ or $\star < L$ should be $\star \leq r$ or $\star \leq L$. That's a huge mistake!
  • There's no need to take the absolute value of $L$. It just complicates matters.
  • It's wasteful to say "subsequence" and "eventually" in the same sentence.
  • So many double negations!
  • It's $\limsup a_n$, not $limsup a_n$.
  • "We claim [X]. … Thus, the claim is established and it follows that [X]." Oh, X follows from itself? How helpful!

Edit: In fact, replacing a few "$<$" with "$\leq$" still won't fix the proof. The first sentence of the last paragraph is wrong either way. It's entirely possible that $\limsup a_n = L$ but $a_n > L$ for all $n$. Just consider $a_n = 1/n$ and $L = 0$.


I would read it so: there must be an $N_1$ such that for all $n\geq N_1, b_n<r-|L|.$ This is because otherwise, for each $N_1\in\Bbb{N}$, we could find a $b_{n_{N_1}}$ with $b_{n_{N_1}}\geq r-|L|$. Then the sequence $b_{n_{N_1}}$ would show that $\lim\sup_{n\to\infty} b_n\geq r$.


I suggest you rather recall that $\limsup x_n<x$ if and only if $x_n<x$ for almost all $n$. Now if $c$ is any number with $c>\limsup a_n+\limsup b_n$ show that you can write $c=a+b$ with $a>\limsup a_n$ and $b>\limsup b_n$, conclude $a_n+b_n<a+b=c$ for almost all $n$ and you are done. Only the case with one limsup being $+\infty$ would need some special care, but then nothing is to be shown.


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