Is it allowed to write $\min\{\delta_k\mid\forall k\in\Bbb N\}$ when nothing is known about the $\delta s$ I encounter this case many times in calculus problems, but I'm never really sure it is legal. 
The question is, can I write  $\min\{\delta_1, \delta_2, \dots\}$ when nothing is known about the $\delta$'s?
If there is a finite number of $\delta s$, all is clear.
But if there is an infinite number of $\delta$'s, how can I make sure the minimum exists?
For example, if I choose $\delta_k=\frac 1 k$, then  the expression $\min\{\delta_k\mid\forall k\in\Bbb N\}$ has no meaning in my view.
 A: Indeed,
$$\min \bigg\{ \frac{1}{k} \ \bigg| \ k \in \Bbb N\bigg\}$$
is a wrong notation.  The $\min$ should be replaced by $\inf$. So yes you are right, without any further information aboute $\{\delta_i\mid i\in\Bbb N\}$ taking the minimum over this set might be a wrong step in general.
A: You are absolutely correct.  In general, one should use the so-called infimum instead of minimum when dealing with an infinite set.  
Oftentimes, infinite collections of real numbers will not have a minimum element, as your example demonstrates.  However, an infinite collection of real numbers that is bounded below will always have an infimum.  The downside is that the infimum might not belong to your original set.  For example:
$$\inf \left( \left\{ \frac{1}{k} \mid k \in \mathbb{N} \right\} \right) = 0$$
However, if a set $S$ is closed and bounded below, then you can guarantee that $\inf(S) \in S$.
A: You are right. If nothing is known about the set of $\delta$'s, the minimum of that set may not exist. However, a similar value is the infimum, or greatest lower bound of the set, which always exists in the reals.
A: Usual sufficient conditions for $A\subset \Bbb R$ to $\min A$ exist (it is not an exhaustive list, and the items are not mutually incompatible):


*

*$A$ is finite.

*$A$ is compact.

*$A$ is closed and bounded below.

*$A$ is an increasing sequence.

*$A$ is a subset of $\Bbb N$, or bounded below subset of $\Bbb Z$.


$\cdots$
