0

I have not studied mathematics, so please be patient with me.

Does it make sense to take the minimum over an uncountable set? In my opinion, if the set is closed in this minimum direction, then it seems to be valid to take the minimum. However, if it is not, then one should take the infimum?

Consider the following example:

$$\min \{ t \mid t \in [0,1]\}$$

Even though the interval $I=[0,1]$ is uncountable, the minimum exists and is $0$, isn't it? This is because this interval is closed at $0$.

However, writing

$$\min \{t \mid t \in (0,1]\}$$

does not really make sense, because there does not exist a minimum. The infimum, however, would again be 0.

Is this correct so far?

So what confuses me is in the context of jump times in stochastic processes. There we define a jump time $S_t$ to be

$S_t= \inf \{ t > S_{t-1} \mid N_t > N_{t-1} \}$,

where $N_t$ counts the number of jumps up to time $t$.

$N_t$ is a integer-valued, right-continuous function. So why do we have to take the infimum and not the minimum here? When it is right-continuous, then we have closed intervals and this value $\min \{ t > S_{t-1} \mid N_t > N_{t-1} \}$ should be well-defined, or am I wrong?

It would be really nice of you if you could point out when we need the infimum and when the minimum and why for jump times we need the infimum.

Thank you very much.

user136457
  • 2,560
  • 1
  • 22
  • 41

1 Answers1

1

Infimum is always defined (if the set is bounded below and nonempty), while a minimum may not exist for such sets, and when it does it is always equal to the infimum. So in many ways the infimum is just a generalization of the minimum, and it is easier to just deal exclusively with $\inf S$; you can write "the minimum exists" in this notation easily as "$\inf S\in S$", so nothing is lost. In your example minimum would also work, but it is no harm to use infimum instead, and indeed you can just use infimum in place of minimum everywhere and you won't change the meaning of anything.

Mario Carneiro
  • 27,438