set notation, find min

Question

I have the following problem:

I have two set of operators $S_1$, $S_2$. Each operator has a cost, which we find it by $Cost(o)$, where $o$ is an operator. Now I need to find the cheapest operator $op$ from the set $S_1\cup S_2$. Is this correctly expressed: $$ op = min_{cost}\{o | o \in S_1 \cup S_2 \} $$ I need to know if I'm expressing it rightly.

Thank you in advance.

Answer 1

Notice that what you express by $\min_f$ is not uniquely determined. There might be many operators which achieve the same minimal cost. This is why such a notation is not perfectly meaningful... but I think it is clear what you mean.

Usually the $\mathop{argmin}$ operator is used with the same meaning: $$ op = \mathop{argmin} \{\mathrm{cost}(o) \colon o \in S_1 \cup S_2\} $$ i.e. the element $o$ where a minimum of a function is achieved.

To be perfectly formal, one should say $$ op \in \mathrm{cost}^{-1}(\{\min \mathrm{cost}(S_1\cup S_2)\}) $$ which has the advantage of making it clear that there is no unique minimizer, but is much less clear to read if one is not familiar with the image and preimage of a set by means of a function.

To be formal and more clear one would define the set of minimizers as $$ \mathcal M = \{o\in S_1\cup S_2 \colon \mathrm{cost}(o)\le \mathrm{cost}(p),\ \forall p \in S_1\cup S_2\} $$ and write $$ op \in \mathcal M. $$

Answer 2

Assuming the minimum exists we can put $$ C_m = \min \{ \operatorname{Cost}(O) \mid O \in S_1 \cup S_2 \} $$ as the minimum cost, and then assuming your $\operatorname{Cost}$ function is bijective you could set $O_m = \operatorname{Cost}^{-1}(C_m)$ to be your minimizing operator.

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The issue I have with your expression is you seem to be minimizing the operators themselves, and in order to do this you need some sort of ordering on the operators (which of course you could define by using the $\operatorname{Cost}$ function, but what I did seems a bit simpler).