# Why does \$sup {x_ky_k}lesup x_kcdot sup y_k\$? Code Answer

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Let \$x_n\$ and \$y_n\$ both be bounded sequences. In the answers proving \$limsup (x_n cdot y_n)lelimsup x_ncdot limsup y_n\$, it is said that, in particular,\$forall n in Bbb{N}\$ \$sup_{kge n} {x_ky_k}lesup_{kge n} x_kcdot sup_{kge n} y_k\$. Why is that? Is this a factclaimcorollary? Something trivial or just something one remembers?

I think you probably mean this in the context of a positive sequence, since \$x_{2k}=-1, x_{2k+1}=-2\$ and \$y_{2k}=-2, y_{2k+1}=-1\$ gives a counterexample otherwise to the fact you quote that \$2=sup_{kge n}x_ky_k> sup_{kge n}x_kcdotsup_{kge n}y_k=1\$.

However, in the positive case we notice that we have an inclusion:

\$\$A={x_ky_k}_ksubseteq {x_my_n}_{m,n}=B\$\$

Now you can prove that \$sup B=(sup x_m)(sup y_n)\$ directly since clearly we just choose subsequences \$n_j\$ and \$m_ell\$ so that

\$\$begin{cases}x_{n_j}to sup x_n\y_{m_ell}to sup y_mend{cases}.\$\$

Then noting \$x_ny_mle (sup x_n)y_mle sup x_ncdot sup y_m\$–which is where we use positivity–you get the result on \$B\$.

Since \$Asubseteq B\$ we have that \$sup Ale sup B\$ by monotonicity of the sup function.

Note that even the case of \$limsup\$ requires the positivity, since my example from the start has that the limsup and sup are equal on \$x_k\$ and \$y_k\$.