Liminf of the maximum of two iid sequences 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!
 A: take a k so that  $P(|X_i | < k, | Y_i | < k ) > \epsilon > 0$.  They must exist because any k for $P(|X| < k) > \frac 3 4 $ and same for Y works,  Let $A_i = \{ |X_i | < k, | Y_i | < k  \}$  By borel cantelli $A_i$  happens infinitely often, and so the liminf is < k
sorry, had misinterpreted dependence structure.  In that case pick any k for which $P( |X_i | < k \} >\frac 12 $ and same for $Y_i$,  Then the density of $\{n> N_0: |X_i | < k \}$ is $ >  \frac 12$ and similarly for Y.  It is easy to argue that therefore they can't be disjoint, and there is and $ n > N_0$ which is in both.
A: This answer contains the details to the answer provided by Michael, all credits go to him.
Choose $M>0$ such that $P(|X_t|<M)>1/2$ and $P(|Y_t|<M)>1/2$. Define $A_t = 1$ if $|X_t|<M$ and 0 otherwise. Similarly, define $B_t = 1$ if $|Y_t|<M$ and 0 otherwise. Then $\{A_t\}_{t\ge1}$ and $\{B_t\}_{t\ge1}$ are iid sequences again. Therefore,
$$
\frac{1}{n}\sum_{t=1}^{n}A_t + B_t \rightarrow E(A) + E(B) = P(|X|<M) + P(|Y|<M) > 1.
$$
We conclude that there must be infinitely many $t$ such that $A_t=B_t=1$.
