Vrftsjtryk

Let ${x_n}$ and ${y_n}$ denote two bounded sequences. Prove that:
$$
lim inf x_n + lim sup y_n le lim sup(x_n + y_n) \
$$

We know that both $x_n$ and $y_n$ are bounded hence is their sum:
$$
m le x_n + y_n < M
$$

Using that fact we may choose a subsequence in order to satisfy the following:
$$
lim(x_{n_k} + y_{n_k}) = limsup(x_n + y_n) tag1
$$

Since $x_{n_k}$ is bounded (as far as $x_n$ is) lets choose a convergent subsequence with indices $n^prime_k ge n_k$ such that:
$$
exists lim x_{n^prime_k}
$$

Now consider a sequence $y_{n^prime_k}$ (note the index is $n^prime_k$), since it is bounded we may choose a convergent subsequence from $y_{n^prime_k}$ with indices $n^{primeprime}_k ge n^prime_k$ such that:
$$
existslim y_{n^{primeprime}_k}
$$

Since ${x_{n^{primeprime}_k}}$ is a subsequence of ${x_{n^prime_k}}$ it is convergent to the same limit. Also ${y_{n^{primeprime}_k}}$ is a subsequence of ${y_{n^prime_k}}$ and we've chosen ${y_{n^{primeprime}_k}}$ to be convergent. Based on that and on $(1)$ we may write:
$$
lim(x_{n^{primeprime}_k} + y_{n^{primeprime}_k}) = lim(x_{n_k} + y_{n_k}) = limsup (x_n + y_n) tag 2
$$

By definition of limsup and liminf:
$$
lim x_{n^{primeprime}_k} ge liminf x_n \
lim y_{n^{primeprime}_k} le limsup y_n
$$

Or (multiply second inequality by $-1$):
$$
lim x_{n^{primeprime}_k} ge liminf x_n \
-lim y_{n^{primeprime}_k} ge -limsup y_n
$$

Subtract the inequalities:
$$
lim x_{n^{primeprime}_k} + lim y_{n^{primeprime}_k} ge liminf x_n + limsup y_n tag3
$$

Limit of sum is just a sum of limits so:
$$
lim(x_{n^{primeprime}_k} + y_{n^{primeprime}_k}) =lim x_{n^{primeprime}_k} + lim y_{n^{primeprime}_k}
$$

So now using $(2)$ and $(3)$ we conclude that:
$$
lim sup(x_n + y_n) ge liminf x_n + limsup y_n
$$

Is this argument enough to consider the proof complete?

The inequality you make after "Subtract the inequalities" needs subtraction instead of addition. If you can find another way to create that inequality, then you are fine.

However, there is a quicker way to answer the problem. By definition of infimum, we have that $displaystyleinf_{kgeq n} x_k leq x_j$ for all $j geq n$. Then we can add $y_j$ to both sides to obtain $$displaystyleinf_{kgeq n} x_k + y_j leq x_j + y_j.$$

Taking the supremum of both sides leads us to
$$displaystyleinf_{kgeq n} x_k + sup_{j geq n} y_j leq sup_{j geq n} (x_j + y_j). $$
Lastly, taking the limit as $n to infty$ yields the result.

Notice that, in the solution, we never used the fact that the sequences are bounded. Thus, this result is actually stronger and can be proved for any real sequences (provided that none of the sums become $infty - infty$).