Let $\{X_t\}_{t\ge1}$ and $\{Y_t\}_{t\ge1}$ be two iid sequences of random variables that have full support. That is, if $A\subseteq\mathbb{R}$ has positive Lebesgue measure, then $P(X\in A) >0$ and $P(Y\in A)>0$. For such a sequence we almost surely have for any $M>0$ that $|X_t|<M$ for infinitely many $t$. That means that with probability one $$ \liminf_{t\rightarrow\infty} |X_t| = 0. $$
Question: Is it possible to find a dependence structure between $\{X_t\}_{t\ge1}$ and $\{Y_t\}_{t\ge1}$ such that $$ P\left(\liminf_{t\rightarrow\infty} \,\,\max\{|X_t|,|Y_t|\} = \infty\right) > 0, $$ that is, with positive probability there exists no $M>0$ such that $\max\{|X_t|,|Y_t|\}<M$ for infinitely many $t$.
Thoughts: I believe this is not possible, because for any subsequence of the $|Y_t|$ that is bounded we need that same subsequence of the $|X_t|$ to go to infinity. Let $t_1,t_2,\ldots$ denote the stochastic times that $|Y_{t_k}|\le M$. Then we must have $\lim_{k\rightarrow\infty}|X_{t_k}| = \infty$. Moreover, $$ P(|X_t|\ge M \,\mid\,|Y_t|< M) = \frac{P(|X_t|\ge M \,;\,|Y_t|< M)}{P(|Y_t|< M)} \le \frac{P(|X_t|\ge M)}{P(|Y_t|< M)} $$ can be made arbitrarily small by increasing $M$. I feel I should somehow be able to combine these results to get the desired answer.
Any help is highly appreciated. Thank you in advance!