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Let $(E,d)$ be a complete separable metric space and $\mathcal X$ be a family of càdlàg functions $E\to[0,\infty)$. Consider the following claim:

  1. For all $t\in[0,\infty)\cap\mathbb Q$, there is a compact $K\subseteq E$ with $$x(t)\in K\;\;\;\text{for all }x\in\mathcal X\tag1.$$
  2. For all $m\in\mathbb N$, $$\left\{x(t):t\in[0,m]\text{ and }x\in\mathcal X\right\}\tag2$$ is relatively compact.

How can we show that 1. and 2. are equivalent?

Now, if $E=\mathbb R$, consider

  1. For all $m\in\mathbb N$, $$\sup_{x\in\mathcal X}\sup_{t\in[0,\:m]}|x(t)|<\infty\tag3.$$

    How can we show that 1. and 3. are equivalent?

Actually, this should be easy, but I'm not sure, especially for 1. and 2., how to do it. Is separability really needed?

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I don't know why you say this should be easy since one of the implications isn't even true.

First, $(2)$ and $(3)$ are clearly equivalent in the case $E = \mathbb{R}$, this is just Heine-Borel. It is also clear that $(2)$ implies $(1)$, even with some uniformity in how you choose $K$.

However $(1)$ implies $(2)$ is false, even in the case $E = \mathbb{R}$. For example, fix a bump function $\phi$ with height $1$ supported in $(-1,1)$. For each $n \geq 1$, define $$\phi_n(\cdot) = n \phi \bigg((n+1)^2(\cdot - 1 + \frac1n)\bigg)$$ The important properties of this construction are that each $\phi_n$ is a bump function with height $n$ and for $n \neq m$, $\operatorname{supp} \phi_n \cap \operatorname{supp} \phi_m = \emptyset$. In particular, at a fixed time $t$, $\{\phi_n(t): n \geq 1\}$ contains just $2$ points and so condition $(1)$ is satisfied.

However, condition $(2)$ fails with $m = 1$, since $\{n : n \geq 1\} \subseteq \{\phi_n(t): t \in [0,1], n \geq 1\}$ and so the right hand set is not relatively compact by Heine-Borel.

Since $(2)$ and $(3)$ are equivalent in this case, this also shows that $(1)$ does not imply $(3)$ when $E = \mathbb{R}$.

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