Let $x_{k}(t)>c>0$ be a bounded sequence for all $t∈(0,1)$ and all $k>0$. Here $t$ is a real parameter never equal zero and $c$ is a fixed real number. Then we have $$\liminf_{k→∞}x_{k}(t)=θ(t)>0$$ Naturally, the $\liminf$ depends on the parameter $t∈(0,1)$.
Does the real number $\sup_{t∈(0,1)}θ(t)$ exist?
I am a little bit confused by the term bounded here. For me bounded means bounded below and above. If this is the case then there is some $d \in \mathbb{R}$ with $$ 0 < c < x_k(t) < d \text{ for all } t \in (0,1), k > 0.$$ This would imply that $c \leq \theta(t) \leq d$ for all $t \in (0,1)$. In other words $\theta(t)$ would then be a bounded sequence and for such a sequence the supremum exists.
When bounded does only mean bounded below then there is an answer to this question as well. For every $t \in (0,1)$ consider the sequence $x_k(t)$ with $$x_k(t) = c+1/t \quad \forall k>0.$$ We have $\theta(t) = c+1/t$ for all $t \in (0,1)$ and for this sequence the supremum does not exist.