Let's start by analyzing each option step-by-step to determine which one is not true.
### Option A: [tex]\( s_n \cup s_{2n} = s_{2n} \)[/tex]
Set [tex]\( s_n = \{100, 101, \ldots, 100 + n\} \)[/tex] and [tex]\( s_{2n} = \{100, 101, \ldots, 100 + 2n\} \)[/tex].
The union [tex]\( s_n \cup s_{2n} \)[/tex] combines all elements from both sets:
- Elements of [tex]\( s_n \)[/tex] are from 100 to [tex]\( 100 + n \)[/tex].
- Elements of [tex]\( s_{2n} \)[/tex] are from 100 to [tex]\( 100 + 2n \)[/tex].
Since [tex]\( 100 + 2n \)[/tex] is always greater than [tex]\( 100 + n \)[/tex], the union of these sets will be exactly [tex]\( s_{2n} \)[/tex], i.e., [tex]\( s_n \cup s_{2n} = s_{2n} \)[/tex].
Thus, Option A is true.
### Option B: If [tex]\( n > m \)[/tex], then [tex]\( s_n \cap s_m = s_n \)[/tex]
Suppose [tex]\( n > m \)[/tex]. Then:
- [tex]\( s_n = \{100, 101, \ldots, 100 + n \} \)[/tex]
- [tex]\( s_m = \{100, 101, \ldots, 100 + m \} \)[/tex]
Since [tex]\( n > m \)[/tex], all elements of [tex]\( s_m \)[/tex] are also in [tex]\( s_n \)[/tex]. Therefore, the intersection [tex]\( s_n \cap s_m \)[/tex] contains exactly the elements of [tex]\( s_m \)[/tex] only, not [tex]\( s_n \)[/tex].
Thus, Option B is false.
### Option C: [tex]\( s_{1000} = \{100, 101, \ldots, 1100\} \)[/tex]
- [tex]\( s_{1000} = \{100, 101, \ldots, 100 + 1000\} = \{100, 101, \ldots, 1100\} \)[/tex]
This statement is directly equal to the definition of [tex]\( s_{1000} \)[/tex].
Thus, Option C is true.
### Option D: [tex]\( s_{60} - s_{30} = \{131, 132, \ldots, 160\} \)[/tex]
- [tex]\( s_{60} = \{100, 101, \ldots, 160\} \)[/tex]
- [tex]\( s_{30} = \{100, 101, \ldots, 130\} \)[/tex]
The difference [tex]\( s_{60} - s_{30} \)[/tex] removes elements that are in [tex]\( s_{30} \)[/tex] from [tex]\( s_{60} \)[/tex], leaving:
- Elements in [tex]\( s_{60} \)[/tex] but not in [tex]\( s_{30} \)[/tex] are [tex]\( 131, 132, \ldots, 160 \)[/tex].
Thus, Option D is true.
### Conclusion
By analyzing the provided options, the correct answer (the statement that is not true) is:
[tex]\[ \boxed{\text{B}} \][/tex]
