Does the real number $\sup_{t∈(0,1)}θ(t)$ exist

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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?

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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.

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Do you know bounded and uniformly bounded?

We call $\{f_n(x)\}$ is bounded when $$\forall n,\,\exists M(n),\,s.t.|f_n(x)|<M(n)$$

And we call $\{f_n(x)\}$ is uniformly bounded (for $n$) when $$\exists M,\,s.t.\forall n,\,|f_n(x)|<M$$

Example:

{$\sin(nx)$}is uniformly bounded (for n) , $M=1$.

{$n\sin(nx)$}is bounded, $M(n)=n$,but isn’t uniformly bounded.

So when the $x_k(t)$ is uniformly bounded (for t), sup exists. Otherwise, no.

Now $x_k(t)$ is just bounded, we let $x_k(t)=k\sin\left(\frac{t\pi}{2}\right)+1$, now $\forall k$, $x_k(t)$ is bounded because $k\geqslant x_k(t)>\frac{1}{2}>0$, and $c=\frac 12$.

But $\forall t\in(0,1)$, $$\limsup_{k→∞}x_{k}(t)=\lim_{k→∞}k \sin\left(\frac{t\pi}{2}\right)+1=+\infty$$ So the $\sup :s(t)$ isn’t exists.