Consider a bounded sequence $x_{n}$ of real numbers. I want to proof that the $limsup$ of a subsequence $x_{kn}$ with $k \in \mathbb{R}$ is smaller or equal to the $limsup$ of the mainsequence $x_{n}$ or in symbols:
$\limsup_{n\to\infty} x_{kn} \leq \limsup_{n\to\infty} x_{n}$.
I can see why this is correct, but don't know how to prove it. I think it is because $x_{kn}$ is a subsequence so if the mainsequence converges, so will the subsequence do, but I can't see how to fix it with the suprema.
Thanks in advance.
If this weren't the case, i.e. \begin{align*} \limsup_{k\to\infty} x_{n_k} > \limsup_{n\to\infty} x_n \doteq c, \end{align*} then for any $N$, we can find an $m(N) > N$ such that $x_{n_{m(N)}} > c+\varepsilon$. Since each $x_{n_{m(N)}}$ is a member of the original sequence, we have constructed a further subsequence $\{x_{n_{m(N)}}\}_{N=1}^{\infty}$ whose limit is strictly greater than $c$ (which by definition is the supremum of all subsequential limits).